Eigenvalues and eigenvectors

Solution 6.6.1

(See Exercise 6.1.) To decide whether the matrix \(A = \left ( \begin{array}{ccc} 2 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & -1 & 2 \end{array} \right )\) is diagonalizable, we follow Method 6.15.

We first compute the characteristic polynomial and eigenvalues. We compute \(\chi_A(t) = \det(A - t \cdot \mathrm{id}_3)\), cf. Definition and Lemma 6.5. Expanding the determinant of

\[ A - t \cdot \mathrm{id}_3 = \left ( \begin{array}{ccc} 2-t & 1 & 1 \\ 0 & 1-t & 0 \\ 1 & -1 & 2-t \end{array} \right ), \]

by developing along the second row (Proposition 5.23), we get

\[ \begin{align*} \chi_A(t) & = (1-t) \det \left ( \begin{array}{cc} 2-t & 1 \\ 1 & 2-t \end{array} \right ) \\ & = (1-t)\bigl[(2-t)^2 - 1\bigr] \\ & = (1-t)(3 - 4t + t^2) \\ & = (1-t)(t-1)(t-3) \\ & = -(t-1)^2(t-3). \end{align*} \]

The eigenvalues are \(\lambda_1 = 1\) (with algebraic multiplicity 2) and \(\lambda_2 = 3\) (with algebraic multiplicity 1).

We now compute the eigenspaces. For \(\lambda_2 = 3\): we solve \((A - 3\,\mathrm{id}_3)v = 0\):

\[ A - 3\,\mathrm{id}_3 = \left ( \begin{array}{ccc} -1 & 1 & 1 \\ 0 & -2 & 0 \\ 1 & -1 & -1 \end{array} \right ) \leadsto \left ( \begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right ). \]

So \(E_3 = \ker(A - 3\,\mathrm{id}_3) = L\bigl((1, 0, 1)\bigr)\), which has dimension 1.

For \(\lambda_1 = 1\): we solve \((A - \mathrm{id}_3)v = 0\):

\[ A - \mathrm{id}_3 = \left ( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 1 & -1 & 1 \end{array} \right ) \leadsto \left ( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right ). \]

So \(E_1 = \ker(A - \mathrm{id}_3) = L\bigl((-1, 0, 1)\bigr)\). Hence the rank is 2, so \(\dim E_1 = 1\).

Finally, we check diagonalizability. By Method 6.15, \(A\) is diagonalizable if and only if \(\dim E_1 + \dim E_3 = 3\). We have \(\dim E_1 = 1\) and \(\dim E_3 = 1\), so \(\dim E_1 + \dim E_3 = 2 \ne 3\). Therefore, \(A\) is not diagonalizable.

Indeed, the eigenvalue \(\lambda_1 = 1\) has algebraic multiplicity 2 but its eigenspace has dimension only 1, which is the obstruction to diagonalizability, cf. also Lemma 6.18.

Solution 6.6.2

(See Exercise 6.2.) For \(A = \left ( \begin{array}{cc} a & b \\ c & d \end{array} \right )\) we have \(A - t\,\mathrm{id}_2 = \left ( \begin{array}{cc} a-t & b \\ c & d-t \end{array} \right )\).

By definition (Definition and Lemma 6.5):

\[ \begin{align*} \chi_A(t) & = \det \left ( \begin{array}{cc} a-t & b \\ c & d-t \end{array} \right ) \\ & = (a-t)(d-t) - bc \\ & = t^2 - (a+d)t + (ad-bc) \\ & = t^2 - \mathrm{tr}(A) t + \det A, \end{align*} \]

where \(\mathrm{tr}(A)=a+d\) is the trace (see Exercise 4.24).

The eigenvalues are the roots of \(\chi_A(t)=0\), hence

\[ t^2 - (a+d)t + (ad-bc) = 0. \]

By the quadratic formula,

\[ \lambda_{1,2} = \frac{a+d}{2} \pm \sqrt{\left(\frac{a+d}{2}\right)^2 - (ad-bc)} = \frac{a+d}{2} \pm \sqrt{\frac{(a-d)^2}{4}+bc}. \]

Solution 6.6.3

(See Exercise 6.3.) We compute \(\chi_A(t)\), eigenvalues, eigenspaces, and diagonalizability for each matrix, following Definition and Lemma 6.5 and Method 6.15.

1. \(A = \left ( \begin{array}{cc} 3 & 5 \\ 1 & -1 \end{array} \right )\).

\[ \begin{align*} \chi_A(t) & = \det \left ( \begin{array}{cc} 3-t & 5 \\ 1 & -1-t \end{array} \right ) = (3-t)(-1-t)-5 \\ & = t^2-2t-8 = (t-4)(t+2). \end{align*} \]

So the eigenvalues of $A$ are $4$ and $-2$. Since these are two distinct eigenvalues, $A$ is diagonalizable by <a href="../eigenvalues/#cor-diag-max" data-reference-type="ref+Label" data-reference="cor:diag-max">Corollary 6.16</a>. One computes the eigenspaces and thus an eigenbasis for example using Gaussian elimination. For $\lambda=4$, $A-4\,\mathrm{id}_2 = \left ( \begin{array}{cc} -1 & 5 \\ 1 & -5 \end{array} \right ) \leadsto x=5y,$ so $E_4 = L((5,1))$.

For $\lambda=-2$, $A+2\,\mathrm{id}_2 = \left ( \begin{array}{cc} 5 & 5 \\ 1 & 1 \end{array} \right ) \leadsto x=-y,$ so $E_{-2} = L((-1,1))$. An eigenbasis for $A$ is $\bigl((5,1),(-1,1)\bigr)$ (cf. <a href="../eigenvalues/#def-eigenbasis" data-reference-type="ref+Label" data-reference="def:eigenbasis">Definition 6.17</a>).

1. For the given matrix \(A\), we get

\[ \begin{align*} \chi_A(t) & = \det \left ( \begin{array}{ccc} -t & 1 & 0 \\ 3 & -t & 1 \\ 2 & 0 & -t \end{array} \right ) \\ & = (-t)\det\left ( \begin{array}{cc} -t & 1 \\ 0 & -t \end{array} \right ) - \det\left ( \begin{array}{cc} 3 & 1 \\ 2 & -t \end{array} \right ) \\ & = -t^3 + 3t + 2 = -(t-2)(t+1)^2. \end{align*} \]

The eigenvalues are $2$ and $-1$ (the latter with algebraic multiplicity 2). Since there are only 2 (not 3) distinct eigenvalues We cannot use <a href="../eigenvalues/#cor-diag-max" data-reference-type="ref+Label" data-reference="cor:diag-max">Corollary 6.16</a> to decide diagonalizability, so we need to compute the eigenspaces. For $\lambda=2$:

\[ A-2\,\mathrm{id}_3 = \left ( \begin{array}{ccc} -2 & 1 & 0 \\ 3 & -2 & 1 \\ 2 & 0 & -2 \end{array} \right ) \leadsto E_2 = L((1,2,1)). \]

For $\lambda=-1$:

\[ A+\mathrm{id}_3 = \left ( \begin{array}{ccc} 1 & 1 & 0 \\ 3 & 1 & 1 \\ 2 & 0 & 1 \end{array} \right ) \leadsto E_{-1} = L((-1,1,2)). \]

Hence $\dim E_2 + \dim E_{-1} = 1+1=2<3$, so $A$ is not diagonalizable; i.e., does not have an eigenbasis.

1. The matrix \(A = \left ( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & \frac 32 \\ 0 & 0 & 1 \end{array} \right )\) is an upper triangular matrix. Thus the eigenvalues are diagonal entries: \(1,0,1\). Thus

\[ \chi_A(t) = (1-t)^2(-t). \]

For $\lambda=1$:

\[ A-\mathrm{id}_3 = \left ( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & -1 & \frac 32 \\ 0 & 0 & 0 \end{array} \right ), \]

so $y=\frac 32 z$, with $x,z$ free. Therefore

\[ E_1 = L((1,0,0),(0,3,2)), \quad \dim E_1=2. \]

For $\lambda=0$ we solve $Av=0$, which gives $x=0$ and $z=0$, so

\[ E_0 = L((0,1,0)), \quad \dim E_0=1. \]

Hence $\dim E_1 + \dim E_0 = 3$, so $A$ is diagonalizable. An eigenbasis is $\bigl((1,0,0),(0,3,2),(0,1,0)\bigr).$

1. The matrix \(A = \left ( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right )\). is again upper triangular, so the only eigenvalue is \(0\) (algebraic multiplicity 3), and

\[ \chi_A(t) = (-t)^3. \]

The eigenspace is $E_0=\ker A$. From

\[ A\left ( \begin{array}{c} x \\ y \\ z \end{array} \right ) = \left ( \begin{array}{c} y \\ z \\ 0 \end{array} \right ) = 0, \]

we get $y=z=0$, $x$ free. So

\[ E_0 = L((1,0,0)), \quad \dim E_0=1. \]

Since $\dim E_0=1<3$, $A$ is not diagonalizable, and there is no eigenbasis.

Solution 6.6.4

(See Exercise 6.4.) Consider

\[ A = \left ( \begin{array}{ccc} 1 & 0 & 0 \\ 1 & 1 & 2 \\ 1 & 0 & 1 \end{array} \right ). \]

We compute the characteristic polynomial using Definition and Lemma 6.5:

\[ \begin{align*} \chi_A(t) & = \det(A-t\,\mathrm{id}_3) \\ & = \det \left ( \begin{array}{ccc} 1-t & 0 & 0 \\ 1 & 1-t & 2 \\ 1 & 0 & 1-t \end{array} \right ) \\ & = (1-t)\det \left ( \begin{array}{cc} 1-t & 2 \\ 0 & 1-t \end{array} \right ) \\ & = (1-t)^3. \end{align*} \]

Hence the only eigenvalue is \(\lambda=1\), with algebraic multiplicity \(3\). The eigenspace is

\[ E_1 = \ker(A-\mathrm{id}_3), \]

and

\[ A-\mathrm{id}_3 = \left ( \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 0 & 2 \\ 1 & 0 & 0 \end{array} \right ) \leadsto \left ( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right ). \]

So the equations are \(x=0\) and \(z=0\), while \(y\) is free. Therefore

\[ E_1 = L((0,1,0)), \qquad \dim E_1 = 1. \]

By Method 6.15, diagonalizability would require the sum of dimensions of eigenspaces to be \(3\). Here there is only one eigenspace and its dimension is \(1\), so \(A\) is not diagonalizable. Consequently, there is no basis of \({\bf R}^3\) in which the associated matrix of \(A\) is diagonal.

Solution 6.6.5

(See Exercise 6.5.) The condition \(\ker f = L((1,1,1))\) implies that \(f(1,1,1)=(0,0,0)\), which we can also rewrite as

\[ f((1,1,1)) = 0 \cdot (1,1,1). \]

Thus, this vector is an eigenvector for \(f\), with eigenvalue 0. We therefore have three eigenvectors as follows:

\[ \begin{align*} v_1 = (1,0,1) & \mapsto 2 (1,0,1)\\ v_2 = (2,0,-3) & \mapsto -1 (2,0,-3) \\ v_3 = (1,1,1) & \mapsto 0 (1,1,1). \end{align*} \]

We check that these three vectors form a basis of \({\bf R}^3\) (note that this is therefore an example of an eigenbasis). To this end, we compute the rank of

\[ \left ( \begin{array}{ccc} 1 & 0 & 1 \\ 2 & 0 & -3 \\ 1 & 1 & 1 \end{array} \right ) \leadsto \left ( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 0 & -5 \\ 0 & 1 & 0 \end{array} \right ). \]

This implies that the matrix has rank three, and therefore the three vectors form a basis of \({\bf R}^3\). The matrix of \(f\) with respect to the basis \(\underline v = \{v_1, v_2, v_3\}\) is

\[ \left ( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{array} \right ). \]

In order to compute the matrix of \(f\) with respect to the standard basis \(\underline e = \{e_1, e_2, e_3 \}\), we use the usual diagram:

\[ {\bf R}^3_{\underline e} \xrightarrow[K]{\mathrm {id}} {\bf R}^3_{\underline v} \xrightarrow[\scriptsize {\left ( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{array} \right )}]{f} {\bf R}^3_{\underline v} \xrightarrow[K^{-1}]{\mathrm {id}} {\bf R}^3_{\underline e}. \]

It turns out that \(K^{-1}\) is easier to compute than \(K\). It is given by expressing the \(v_i\) in their coordinates in the standard basis vectors, e.g. \(v_1 \mapsto {\mathrm {id}}(v_1) = (1,0,1) = 1 e_1 + 0 e_2 + 1 e_3\). This implies \(K^{-1} = \left ( \begin{array}{ccc} 1 & 2 & 1 \\ 0 & 0 & 1 \\ 1 & -3 & 1 \end{array} \right )\). We can use this to compute \(K = (K^{-1})^{-1})\), cf.. This inverse (of \(K^{-1}\)) can be computed using Theorem 4.80, which gives \(K = \frac 15 \left ( \begin{array}{ccc} 3 & -5 & 2 \\ 1 & 0 & -1 \\ 0 & 5 & 0 \end{array} \right )\). Then, one computes the product

\[ K^{-1} \left ( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{array} \right ) K = \frac 15 \left ( \begin{array}{ccc} 4 & -10 & 6 \\ 0 & 0 & 0 \\ 9 & -10 & 1 \end{array} \right ). \]

This is the basis of \(f\) with respect to the standard basis.

This is a typical example of the situation that one basis of \({\bf R}^3\) may be more adapted to describing a linear map than another one. An eigenbasis, such as \(v_1, v_2, v_3\) gives a particularly simple matrix.

Solution 6.6.6

(See Exercise 6.6.) For

\[ A_a = \left ( \begin{array}{ccc} a & 0 & 0 \\ a-2 & 1 & 1 \\ 0 & 1 & 1 \end{array} \right ), \]

we compute the characteristic polynomial (Definition and Lemma 6.5):

\[ \begin{align*} \chi_{A_a}(t) &= \det(A_a - t\,\mathrm{id}_3) \\ &= \det \left ( \begin{array}{ccc} a-t & 0 & 0 \\ a-2 & 1-t & 1 \\ 0 & 1 & 1-t \end{array} \right ) \\ &= (a-t)\det \left ( \begin{array}{cc} 1-t & 1 \\ 1 & 1-t \end{array} \right ) \\ &= (a-t)\bigl((1-t)^2-1\bigr) = (a-t)\,t\,(t-2). \end{align*} \]

So the eigenvalues are \(a,0,2\) (with multiplicities depending on \(a\)). We distinguish several cases:

1. If \(a\neq 0,2\), there are three distinct eigenvalues, hence \(A_a\) is diagonalizable (Corollary 6.16).

2. If \(a=0\), then \(A_0 = \left ( \begin{array}{ccc} 0 & 0 & 0 \\ -2 & 1 & 1 \\ 0 & 1 & 1 \end{array} \right ).\) To decide diagonalizability, we note that the eigenspace of \(\lambda=2\) is one-dimensional, since the algebraic multiplicity of \(\lambda=2\) is 1. We now compute the eigenspace for \(\lambda=0\):

\[ A_0 - 0 \,\mathrm{id}_3 = A_0 = \left ( \begin{array}{ccc} 0 & 0 & 0 \\ -2 & 1 & 1 \\ 0 & 1 & 1 \end{array} \right ) \leadsto \left ( \begin{array}{ccc} -2 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array} \right ). \]

This gives $x = \frac{1}{2}y + \frac{1}{2}z$ and $y + z = 0$, so $y = -z$. Thus $x = 0$ and $E_0=L((0,1,-1))$, so $\dim E_0=1$ while the algebraic multiplicity of $0$ is $2$. Therefore $A_0$ is not diagonalizable.

1. If \(a=2\), then

\[ A_2 = \left ( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{array} \right ). \]

This time, the algebraic multiplicity of $\lambda = 0$ is 1, so it remains to compute the eigenspace for $\lambda = 2$:

\[ A_2-2\,\mathrm{id}_3 = \left ( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 1 & -1 \end{array} \right ), \]

so $y=z$ and $x$ is free; hence

\[ E_2 = L((1,0,0),(0,1,1)), \qquad \dim E_2=2. \]

For $\lambda=0$, one gets $E_0=L((0,1,-1))$, thus $\dim E_0=1$. So $\dim E_2+\dim E_0=3$, and by <a href="../eigenvalues/#met-diagonalizability" data-reference-type="ref+Label" data-reference="met:diagonalizability">Method 6.15</a>, $A_2$ is diagonalizable.

In conclusion, \(A_a\) is diagonalizable precisely if \(a \ne 0\).

Solution 6.6.7

(See Exercise 6.7.) We have \(\det (A_a - t {\mathrm {id}}_3) = \det \left ( \begin{array}{ccc} 4-t & 0 & 4 \\ a & 2-t & a \\ -2 & 0 & -2-t \end{array} \right )\). It is convenient to develop along the second column (Proposition 5.23), since it has two zeros; other developments are possible and give the same result, but are more tedious to perform. This gives

\[ \begin{align*} \det (A_a - t {\mathrm {id}}_3) & = (2-t) \det \left ( \begin{array}{cc} 4-t & 4 \\ -2 & -2-t \end{array} \right ) \\ & = (2-t)\bigl[(4-t)(-2-t) + 8\bigr] \\ & = (2-t)\bigl[-8-4t+2t+t^2 + 8\bigr] \\ & = (2-t)(t^2-2t) \\ & = (2-t) \cdot t(t-2) \\ & = -(t-2)^2 t. \end{align*} \]

The roots of this polynomial, i.e., the eigenvalues are \(2\) and 0 (regardless of the value of \(a\)). The exponent of \(t-2\) in the above polynomial is 2, the one for \(t\) is 1. This implies that

\[ \begin{align*} 1 \le \dim E_0 \le 1 & \ \text{for all } t \in {\bf R} \\ 1 \le \dim E_2 \le 2 & \ \text{for all } t \in {\bf R}. \end{align*} \]

According to Method 6.15, \(A_a\) will be diagonalizable precisely if \(\dim E_2 = 2\). We compute \(E_2\) by bringing \(A_a - 2 {\mathrm {id}}\) into reduced row echelon form:

\[ \begin{align*} \left ( \begin{array}{ccc} 4-2 & 0 & 4 \\ a & 2-2 & a \\ -2 & 0 & -2-2 \end{array} \right ) & = \left ( \begin{array}{ccc} 2 & 0 & 4 \\ a & 0 & a \\ -2 & 0 & -4 \end{array} \right ) \leadsto \left ( \begin{array}{ccc} 2 & 0 & 4 \\ a & 0 & a \\ 0 & 0 & 0 \end{array} \right ) \\ & \leadsto \left ( \begin{array}{ccc} 1 & 0 & 2 \\ a & 0 & a \\ 0 & 0 & 0 \end{array} \right ) \leadsto \left ( \begin{array}{ccc} 1 & 0 & 2 \\ 0 & 0 & a-2a \\ 0 & 0 & 0 \end{array} \right ) \\ & = \left ( \begin{array}{ccc} 1 & 0 & 2 \\ 0 & 0 & -a \\ 0 & 0 & 0 \end{array} \right ). \end{align*} \]

This matrix has rank 1, or equivalently \(\dim E_2 = 2\), if and only if \(a=0\). Thus, the matrix \(A_a\) is diagonalizable precisely if \(a=0\). The second part of the exercise then has only to be done for \(a=0\), i.e. \(A := A_0 = \left ( \begin{array}{ccc} 4 & 0 & 4 \\ 0 & 2 & 0 \\ -2 & 0 & -2 \end{array} \right )\). This can be dealt with as in the previous exercises.

Solution 6.6.8

(See Exercise 6.8.) If \(A = \left ( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 2 & 0 \\ 1 & -1 & 1 \end{array} \right )\) and \(B = \left ( \begin{array}{ccc} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \end{array} \right ).\) represent the same linear map (with respect to different bases), there needs to be a base change matrix \(P\) such that \(B = P^{-1}AP\). This will imply that \(A\) and \(B\) have the same characteristic polynomial, and in particular the same eigenvalues (see Proposition 6.24). We compute

\[ \begin{align*} \chi_A(t) &= \det(A-t\,\mathrm{id}_3) \\ &= \det \left ( \begin{array}{ccc} 1-t & 1 & 1 \\ 0 & 2-t & 0 \\ 1 & -1 & 1-t \end{array} \right ) \\ &= (2-t)\det \left ( \begin{array}{cc} 1-t & 1 \\ 1 & 1-t \end{array} \right ) \\ &= (2-t)\bigl((1-t)^2-1\bigr) = -t(t-2)^2. \end{align*} \]

Hence the eigenvalues are \(0\) and \(2\) (with algebraic multiplicity \(2\) for \(2\)). The matrix \(B\) is upper triangular, so

\[ \chi_B(t) = (2-t)^2(-t) = -t(t-2)^2 = \chi_A(t). \]

Thus \(A\) and \(B\) have the same characteristic polynomial and the same eigenvalues. So, from considering only the characteristic polynomial we cannot conclude that \(A\) and \(B\) do not represent the same linear map with respect to different bases.

We proceed by computing the eigenspaces. In fact, again if \(B = P^{-1}AP\), then the eigenspaces (for any \(\lambda \in \mathbf{R}\)) of \(A\) and \(B\) are the same. We begin by computing the eigenspaces for \(A\). For \(\lambda=0\), we solve \(Av=0\):

\[ \left\{\begin{array}{l} x+y+z=0 \\ 2y=0 \\ x-y+z=0 \end{array}\right. \iff \left\{\begin{array}{l} y=0 \\ x=-z \end{array}\right., \]

so \(E_0(A)=L((1,0,-1))\). For \(\lambda=2\), we solve \((A-2\,\mathrm{id}_3)v=0\):

\[ A-2\,\mathrm{id}_3 = \left ( \begin{array}{ccc} -1 & 1 & 1 \\ 0 & 0 & 0 \\ 1 & -1 & -1 \end{array} \right ), \]

which gives \(-x+y+z=0\), i.e. \(y=x-z\). Therefore

\[ E_2(A) = L((1,1,0),(0,-1,1)), \qquad \dim E_2(A)=2. \]

So

\[ \dim E_0(A)+\dim E_2(A)=1+2=3, \]

and by Method 6.15, \(A\) is diagonalizable.

For \(\lambda=0\):

\[ Bv=0 \iff \left\{\begin{array}{l} 2x+y=0 \\ 2y=0 \end{array}\right. \iff x=y=0, \]

so \(E_0(B)=L((0,0,1))\). For \(\lambda=2\):

\[ B-2\,\mathrm{id}_3 = \left ( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -2 \end{array} \right ), \]

so \(y=0\), \(z=0\), \(x\) free, and

\[ E_2(B)=L((1,0,0)), \qquad \dim E_2(B)=1. \]

Hence

\[ \dim E_0(B)+\dim E_2(B)=1+1=2<3, \]

so \(B\) is not diagonalizable.

In particular, the eigenspaces of \(A\) and \(B\) are not the same, which implies that \(A\) and \(B\) do not represent the same linear map with respect to different bases.

Solution 6.6.9

(See Exercise 6.9.) For

\[ A_t = \left ( \begin{array}{ccc} -1 & 2 & t \\ 2 & 0 & -2 \\ t & -2 & -1 \end{array} \right ), \]

the value \(0\) is an eigenvalue precisely when

\[ \det(A_t)=0 \]

(Theorem 6.8). (Alternatively, one may also compute the rank of \(A_t\) and check when it is \(\le 2\).) By developing along the first row (Proposition 5.23), we compute

\[ \begin{align*} \det(A_t) &= -1\det\left ( \begin{array}{cc} 0 & -2 \\ -2 & -1 \end{array} \right ) -2\det\left ( \begin{array}{cc} 2 & -2 \\ t & -1 \end{array} \right ) +t\det\left ( \begin{array}{cc} 2 & 0 \\ t & -2 \end{array} \right ) \\ &= 4 + (-4t+4) -4t \\ &= 8(1-t). \end{align*} \]

Hence \(0\) is an eigenvalue if and only if \(t=1\).

So we now consider

\[ A_1=\left ( \begin{array}{ccc} -1 & 2 & 1 \\ 2 & 0 & -2 \\ 1 & -2 & -1 \end{array} \right ). \]

Its characteristic polynomial is

\[ \begin{align*} \chi_{A_1}(\lambda) &=\det(A_1-\lambda\,\mathrm{id}_3) = -\lambda(\lambda+4)(\lambda-2). \end{align*} \]

Therefore the eigenvalues are \(\lambda_1=0\), \(\lambda_2=2\), \(\lambda_3=-4.\) For \(\lambda=0\), solve \(A_1v=0\):

\[ \left\{\begin{array}{l} -x+2y+z=0 \\ 2x-2z=0 \end{array}\right. \Rightarrow x=z,\ y=0, \]

so \(E_0=L((1,0,1))\). Similarly, one computes the eigenspaces \(E_2=L((-1,-2,1))\) and \(E_{-4}=L((-1,1,1))\).

Solution 6.6.10

(See Exercise 6.10.) Let

\[ A = \left ( \begin{array}{cc} -4 & 8 \\ 1 & -2 \end{array} \right ), \qquad F : {\mathrm {Mat}}_{2\times 2} \to {\mathrm {Mat}}_{2\times 2},\ X\mapsto AX. \]

This map is indeed linear by Lemma 4.59. Fixing the basis \(\underline E := (E_{11},E_{12},E_{21},E_{22})\), we compute:

\[ \begin{align*} F(E_{11}) &= AE_{11} = \left ( \begin{array}{cc} -4&0\\1&0 \end{array} \right ) = -4E_{11}+E_{21},\\ F(E_{12}) &= AE_{12} = \left ( \begin{array}{cc} 0&-4\\0&1 \end{array} \right ) = -4E_{12}+E_{22},\\ F(E_{21}) &= AE_{21} = \left ( \begin{array}{cc} 8&0\\-2&0 \end{array} \right ) = 8E_{11}-2E_{21},\\ F(E_{22}) &= AE_{22} = \left ( \begin{array}{cc} 0&8\\0&-2 \end{array} \right ) = 8E_{12}-2E_{22}. \end{align*} \]

Therefore the matrix of \(F\) with respect to this basis (in the domain and codomain) is

\[ M := \left ( \begin{array}{cccc} -4 & 0 & 8 & 0 \\ 0 & -4 & 0 & 8 \\ 1 & 0 & -2 & 0 \\ 0 & 1 & 0 & -2 \end{array} \right ). \]

We now compute \(\ker F\) and \(\operatorname{im} F\) by Gaussian elimination on this \(4\times 4\) matrix. Set

\[ M \leadsto \left ( \begin{array}{cccc} 1 & 0 & -2 & 0 \\ 0 & 1 & 0 & -2 \\ 0 & -4 & 0 & 8 \\ -4 & 0 & 8 & 0 \end{array} \right ) \leadsto \left ( \begin{array}{cccc} 1 & 0 & -2 & 0 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right ). \]

Hence, for coordinates \((x_1,x_2,x_3,x_4)\) in the basis \(\underline E=(E_{11},E_{12},E_{21},E_{22})\), the kernel equations are \(x_1-2x_3=0\) and \(x_2-2x_4=0\). So

\[ \ker F =L\bigl((2,0,1,0),(0,2,0,1)\bigr) =L(2E_{11}+E_{21},\ 2E_{12}+E_{22}) =L\left(\left ( \begin{array}{cc} 2 & 0 \\ 1 & 0 \end{array} \right ),\left ( \begin{array}{cc} 0 & 2 \\ 0 & 1 \end{array} \right )\right) . \]

From the same row-reduction we read \(\operatorname{rank} M=2\) and pivot columns \(1,2\). Therefore

\[ \operatorname{im} F =L\bigl(c_1,c_2\bigr) =L\bigl((-4,0,1,0),(0,-4,0,1)\bigr) =L(-4E_{11}+E_{21},\ -4E_{12}+E_{22}) =L\left(\left ( \begin{array}{cc} -4 & 0 \\ 1 & 0 \end{array} \right ),\left ( \begin{array}{cc} 0 & -4 \\ 0 & 1 \end{array} \right )\right), \]

where \(c_1,c_2\) are the first two columns of \(M\).

For eigenvalues and eigenspaces, note that the eigenvector condition \(F(X)=\lambda X\) is equivalent to \(Au=\lambda u\ \text{and}\ Av=\lambda v\). Indeed, this holds since \(A(u \ v) = (Au \ Av)\), for a \(2\times 2\) matrix with columns \(u,v\). So \(u,v\) must be eigenvectors for \(A\) (for some eigenvalue \(\lambda\)). Now

\[ \chi_A(t)=\det(A-t\,\mathrm{id}_2) =\det\left ( \begin{array}{cc} -4-t&8\\1&-2-t \end{array} \right ) =t(t+6), \]

so the eigenvalues of \(A\) are \(0\) and \(-6\). Therefore the eigenvalues of \(F\) are also \(0\) and \(-6\).

For \(\lambda=0\), the eigenspace is exactly \(\ker F\), hence

\[ E_0(F) =L\left(\left ( \begin{array}{cc} 2 & 0 \\ 1 & 0 \end{array} \right ),\left ( \begin{array}{cc} 0 & 2 \\ 0 & 1 \end{array} \right )\right) =L(2E_{11}+E_{21},\ 2E_{12}+E_{22}). \]

For \(\lambda=-6\), first compute

\[ E_{-6}(A)=L\left(\left ( \begin{array}{c} -4\\1 \end{array} \right )\right). \]

Thus both columns of \(X\) must belong to the 1-dimensional subspace of \(\mathbf R^2\) spanned by this vector, so

\[ E_{-6}(F) =L\left(\left ( \begin{array}{cc} -4 & 0 \\ 1 & 0 \end{array} \right ),\left ( \begin{array}{cc} 0 & -4 \\ 0 & 1 \end{array} \right )\right) =L(-4E_{11}+E_{21},\ -4E_{12}+E_{22}). \]

So, at the end of the computation, the eigenspaces of \(F\) are explicitly

\[ E_0(F)=L\left(\left ( \begin{array}{cc} 2 & 0 \\ 1 & 0 \end{array} \right ),\left ( \begin{array}{cc} 0 & 2 \\ 0 & 1 \end{array} \right )\right), \qquad E_{-6}(F)=L\left(\left ( \begin{array}{cc} -4 & 0 \\ 1 & 0 \end{array} \right ),\left ( \begin{array}{cc} 0 & -4 \\ 0 & 1 \end{array} \right )\right). \]

Solution 6.6.11

(See Exercise 6.11.) If \(A\) and \(B\) represent the same map, then \(A = PBP^{-1}\) for some invertible matrix \(P\). By Proposition 5.18, this implies that \(\det A = \det P \det B \det P^{-1} = \det B\). In short, \(A\) and \(B\) have to have the same determinant. This is true: \(\det A = \det B = 9\).

In addition, again by Proposition 5.18, \(A\) and \(B\) have to have the same characteristic polynomial:

\[ \begin{align*} \chi_A(t) & = \det (A - t{\mathrm {id}}) \\ & = \det \left ( \begin{array}{ccc} 1-t & 0 & 2 \\ 0 & 3-t & 0 \\ 0 & 0 & 3-t \end{array} \right ) \\ & = (1-t)(3-t)^2 \\ & = \det (B-t{\mathrm {id}}) = \chi_B(t). \end{align*} \]

Again, this is true.

Finally, the dimensions of the eigenspaces of the eigenvalues (\(1\) and \(3\)) need to be equal. For \(A\), the eigenspace \(E_{1, A}\) for the eigenvalue \(\lambda = 1\) has \(\dim E_{1,A} = 2\) (as one computes!). For \(B\) instead, \(\dim E_{1, B} = 1\). Therefore \(A\) and \(B\) do not represent the same linear map.

Solution 6.6.12

(See Exercise 6.12.) The matrix \(A\) has eigenvalues \(1\) and \(2\). The eigenspaces are computed as \(E_1 = L(1,0,0)\) and \(E_2 = L(1,1,0)\). Their dimensions sum up to 2, which is strictly less than 3, so that \(A\) is not diagonalizable.

The matrix \(A^2\) therefore has \(2^2 = 4\) as an eigenvalue. Since similar matrices have the same eigenvalues (Proposition 6.24), \(A\) is not similar to \(A^2\).

Solution 6.6.13

(See Exercise 6.13.) We first check that \(v_1, v_2, v_3\) are a basis of \({\bf R}^3\). Indeed, one can compute

\[ \det \left ( \begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 2 \end{array} \right ) = 1 \ne 0, \]

so the rank is 3 and the vectors do form a basis.

The condition that \(v_3\) be an eigenvector for the eigenvalue 4 means \(f(v_3) = 4v_3 = (4,4,8)\). Acccording to Proposition 4.39, there is a unique linear map \(f\) whose value on \(v_1, v_2, v_3\) is prescribed.

We now compute \(A\). We have to express \(f(v_i)\) as a linear combination in terms of \(v_1, v_2, v_3\):

\[ \begin{align*} f(v_1) & = (0,0,0) = 0 v_1 + 0v_2 + 0v_3 \\ f(v_2) & = (1,0,3) = a v_1 + bv_2+cv_3 \\ f(v_3) & = (4,4,8) = 0 v_1 + 0v_2 + 4 v_3. \end{align*} \]

We compute \(a,b,c\) above by solving the system

\[ (1,0,3) = (a+b+c, b+c, a+b+2c). \]

Thus \(b=-c\), \(1=a\) and then \(c=2\). Thus

\[ A = \left ( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & -2 & 0 \\ 0 & 2 & 4 \end{array} \right ). \]

In order to compute \(B\) we could use the base change matrix, but it is also possible to compute \(B\) directly. We will express the standard basis vectors \((1,0,0)\) as a linear combination of the \(v_1, v_2, v_3\):

\[ \begin{align*} av_1 +bv_2 +cv_3 & = a(1,0,1) + b(1,1,1)+c(1,1,2) \\ & =(a+b+c,b+c,a+b+2c). \end{align*} \]

Thus, the equation \((1,0,0) = av_1+bv_2 +cv_3\) amounts to the linear system

\[ \begin{align*} 1 & = a+b+c \\ 0 &= b+c \\ 0 &= a+b+2c \end{align*} \]

One solves this: \(a=1\), \(b=1\), \(c=-1\). Similarly, one solves the linear system \((0,1,0) = av_1+bv_2+cv_3\). Its solution is \(a=-1\), \(b=1\), \(c=0\). Finally, for \((0,0,1)=av_1+bv_2+cv_3\) one gets the solution \(a=0\), \(b=-1\), \(c=1\).

Thus, since \(f\) is linear (this is the key point!, cf. Definition 4.1), we have

\[ \begin{align*} f(1,0,0) & = f(v_1+v_2-v_3) \\ & = f(v_1) +f(v_2)-f(v_3) \\ &= (0,0,0)+(1,0,3)-4(1,1,2) \\ &=(-3,-4,-5). \end{align*} \]

Likewise

\[ \begin{align*} f(0,1,0) &= f(-v_1+v_2) = -f(v_1)+f(v_2) = (1,0,3) \\ f(0,0,1)& =f(-v_2+v_3) = -f(v_2)+f(v_3) = (3,4,5) \\ \end{align*} \]

Therefore (writing \(f(1,0,0)\) as the first column etc.), we get

\[ B = \left ( \begin{array}{ccc} -3 & 1 & 3 \\ -4 & 0 & 4 \\ -5 & 3 & 5 \end{array} \right ). \]

The vector \(v_t\) belongs to the image precisely if is a linear combination of the vectors \(f(v_1) = 0\), \(f(v_2)=(1,0,3)\) and \(f(v_3) = (4,4,8)\). This translates into the linear system

\[ \begin{align*} a+3b & = 2\\ 4b&=t\\ 3a+5b&= 5. \end{align*} \]

One solves the first and third equation to \(a = \frac 54\), \(b=\frac 14\). Therefore, the system has a solution precisely if \(t=1\). (Alternatively, one may also solve the linear system \(B \left ( \begin{array}{c} x \\ y \\ z \end{array} \right ) = \left ( \begin{array}{c} 2 \\ t \\ 5 \end{array} \right )\).)

Solution 6.6.14

(See Exercise 6.14.) By Theorem 5.15, \(A\) is invertible precisely iff \(\det A = 0\). We compute the determinant, for example using Sarrus’ rule (Lemma 5.11), as \(\det A = 0 - 24 + 6t-10t -0+30 = 6-4t\). Thus, the condition \(\det A = 6-4t = 0\) amounts to \(t = \frac 32\). The matrix \(A\) is therefore not invertible precisely if \(t = \frac 32\).

We compute the eigenvalues of \(A = \left ( \begin{array}{ccc} 0 & 2 & 2 \\ -3 & -5 & 6 \\ -2 & -2 & 5 \end{array} \right )\) by computing its characteristic polynomial. It is given by

\[ \begin{align*} \chi_A(t) & = t(5+t)(5-t)-24+12+4(-5-t)-12t+6(5-t) \\ & = -t^3 + 3t-2. \end{align*} \]

One zero of this polynomial is \(t = 1\). Dividing the above polynomial by \(t-1\) gives \(-t^2 -t+2\), which has zeroes 1 and \(-2\), respectively. Thus

\[ \chi_A(t) = -(t-1)^2 (t+2). \]

The eigenvalues of \(A\) are therefore \(\lambda = 1\) and \(\lambda = -2\).

We compute the eigenspaces by bringing \(A - \lambda {\mathrm {id}}\) into row echelon form

\[ \begin{align*} A - (-2) {\mathrm {id}} & = \left ( \begin{array}{ccc} 2 & 2 & 2 \\ -3 & -3 & 6 \\ -2 & -2 & 3 \end{array} \right ) \leadsto \left ( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & -2 \\ 2 & 2 & -3 \end{array} \right ) \\ & \leadsto \left ( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{array} \right ) \leadsto \left ( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right ). \end{align*} \]

This matrix has rank 2, and its kernel is thus 1-dimensional. It is spanned by \((1,-1,0)\). Similarly

\[ \begin{align*} A - {\mathrm {id}} & = \left ( \begin{array}{ccc} -1 & 2 & 2 \\ -3 & -3 & 6 \\ -2 & -2 & 4 \end{array} \right ) \leadsto \left ( \begin{array}{ccc} 1 & -2 & -2 \\ 0 & 1 & 0 \\ 0 & 1 & 0 \end{array} \right ) \\ & \leadsto \left ( \begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right ). \end{align*} \]

This again has rank 2, so that the eigenspace \(E_1\) is again 1-dimensional. It is spanned by \((2,0,1)\).

Since \(v =(2,0,a)\) was requested to be an eigenvector, it will be in one of the two eigenspaces. One sees it must lie in \(E_1\), and \((2,0,a)\) lies in \(E_1\) precisely if \(a=1\). Thus \(v=(2,0,1)\), and its eigenvalue is 1.

The matrix \(A\) is not diagonalizable, since \(\dim E_2 + \dim E_1 = 2 < 3\).

The matrix \(A\) is not similar to \(A^2\) since similar matrices have the same determinant. Above we computed \(\det A = 6-4t\), so for \(t = 2\) we have \(\det A = -2\), so that \(\det A^2 = (-2)^2 = 4 \ne \det A\).

These computations can also be performed using any computer algebra software such as Wolfram Alpha:

• https://www.wolframalpha.com/input?i=Characteristic+polynomial+%28%280%2C2%2C2%29%2C%28-3%2C-5%2C6%29%2C%28-2%2C-2%2C5%29%29 (characteristic polynomial)

• https://www.wolframalpha.com/input?i=Ker+%28+%28%280%2C2%2C2%29%2C%28-3%2C-5%2C6%29%2C%28-2%2C-2%2C5%29%29+-+%28%281%2C0%2C0%29%2C%280%2C1%2C0%29%2C%280%2C0%2C1%29%29%29 (eigenspace)

Solution 6.6.15

(See Exercise 6.15.) The kernel of \(A\) is different from \(\{0\}\) precisely if \(A\) is not invertible or, equivalently, if \(\det A = 0\). For example using Sarrus’ rule (Lemma 5.11), we have \(\det A = 2t-6\), so \(t = 3\).

For \(t=3\), we have \(\chi_A(t) = -t^3+6t^2-8t = -t(t^2-6t+8) = -t(t-2)(t-4)\). The eigenvalues are then 0, 2 and 4.

The eigenspaces are \(E_0 = \ker A = L(1,-5,2)\), \(E_2 = \ker (A-2{\mathrm {id}}) = L(1,1,0)\) and \(E_4 = \ker (A-4{\mathrm {id}}) = L(-1,1,2)\). The matrix \(P = \left ( \begin{array}{ccc} 1 & 1 & -1 \\ -5 & 1 & 1 \\ 2 & 0 & 2 \end{array} \right )\) (whose columns are the three eigenvectors comprising an eigenbasis) is then such that \(PAP^{-1} = \left ( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{array} \right )\).

If \(B\) satisfies \(\chi_B(t)=\chi_A(t)=-t(t-2)(t-4)\), its eigenvalues are 0, 2 and 4. These are 3 distinct eigenvalues (i.e., equal to the size of the matrix), so \(B\) is diagonalizable.

Solution 6.6.16

(See Exercise 6.16.) We compute \(\det A = -4t-2\), this is 0 precisely if \(t=-\frac 12\). Thus, \(A\) is non-invertible for \(t=-\frac 12\) and invertible otherwise.

We compute the characteristic polynomial. It is helpful to develop the determinant of \(A - \lambda {\mathrm {id}}\) along the second column (Proposition 5.23), since there are two 0’s in this column, simplifying the formula:

\[ \begin{align*} \det (A-\lambda {\mathrm {id}}) & = (2-\lambda) \det \left ( \begin{array}{cc} 1-\lambda & t \\ 2 & -1-\lambda \end{array} \right ) \\ & = (2-\lambda)(\lambda^2 - 2t - 1). \end{align*} \]

The eigenvalues of \(A\) are therefore \(\lambda_1 = 2\) and \(\lambda_{2/3} = \pm \sqrt{2t+1}\). The latter two eigenvalues are real numbers precisely if \(2t+1 \ge 0\), i.e., if \(t \ge -\frac 12\).

An eigenvalue appears with multiplicity 2 in the above characteristic polynomial precisely if either \(2t+1=0\) (so that \(\lambda_2=\lambda_3\)), i.e., \(t=-\frac 12\), or if \(\sqrt{2t+1} = 2\), i.e., if \(t=\frac 32\).

For \(t=-\frac 12\), we compute the eigenspaces for the eigenvalues \(\lambda_1 = 2\) and \(\lambda_2 = 0\). These are both 1-dimensional, respectively given by \(E_2 = L(0,1,0)\) and \(E_0 = L(2,-3,4)\). The sum of their dimensions is 2, which is less than 3, so \(A\) is not diagonalizable (Method 6.15).