# Determining $\det(\mathbf{A})$ using the characteristic polynomial

Let the 3x3 matrix be $\mathbf{A} = \begin {bmatrix} 3&1&0\\1&3&0\\0&0&1 \end {bmatrix}$.

a) Determine its eigenvalues and eigenvectors.

c) Show the diagonalization of $\mathbf{A}$.

d) Determine $\mathbf{A}^{10}$ without using $\mathbf{A}$.

e) By using the characteristic polynomial of part a), determine $\det(\mathbf{A})$.

a) From the characteristic equation I found eigenvalues as 1, 2 and 4. And from the equation $(\lambda I - A)x = 0$, I found the set of eigenvectors as $$\left\{(x_1, x_2, x_3) \mid x_1 = x_2 = 0, x_3 \neq 0\right\} \cup \left\{(x_1, x_2, x_3) \mid x_1 = -x_2 \neq 0, x_3 = 0 \right\} \cup \left\{ (x_1, x_2, x_3) \mid x_1 = x_2 \neq 0, x_3 = 0\right\}.$$

b) I said "For a set to be orthonormal it must be an orthogonal set and all vectors in the set must have the unit norm. But not all pairs of eigenvectors of $A$ gives zero when we take their inner product. For example $(0,0,1)$ is an eigenvector of $A$, so is $(0,0,2)$, and their inner product gives 2, not 0. Besides, it's obvious that not all eigenvectors have the unit norm. So, the eigenvectors of $A$ does not form an orthonormal set."

I'll skip the solutions of parts c) and d) because the part with which I actually have trouble is part e). I don't know what to do in e). Can somebody help me?

• For any $n\times n$ matrix $A$, the constant coefficient of the characteristic polynomial is $(-1)^n\det(A)$. – Casteels Dec 25 '14 at 12:36
• Also for $b$, while you are right technically, I suspect they might mean to pick one eigenvector for each eigenvalue and show that the set of just these three forms an orthonormal set. – Casteels Dec 25 '14 at 12:47

Running from the beginning, we have that:

$$\chi_{\mathbf{A}}=\begin{vmatrix}3-\lambda & 1 & 0 \\ 1 & 3 - \lambda & 0 \\ 0 & 0 & 1-\lambda\end{vmatrix}=(1-\lambda)(\lambda^{2}-6\lambda+8)$$

Thus we have that: $$\lambda=\{1,2,4\}$$

As you found. Now we can find the eigenvectors for each eigenvalue by solving $(\mathbf{A}-\lambda \mathbf{I})\vec{x}=\vec{0}$, doing this we get the three principle eigenvectors:

$$\vec{v}_{1}=\frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix},\quad \vec{v}_{2}=\frac{1}{\sqrt{2}}\begin{pmatrix}-1 \\ 1 \\ 0\end{pmatrix}, \quad \vec{v}_{3}=\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$$

We note that all of these vectors are normalized, and taking the inner product we have:

$$\vec{v}_{i} \cdot \vec{v}_{j}=\delta_{ij}, \qquad \forall i,j \in \{1,2,3\}$$

Where $\delta_{ij}$ is the Kronecker delta. And thus they form an orthonormal set.

For the next part we note that the spectral theorem implies that:

$$\mathbf{A}=\mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^{-1}$$

Where $\mathbf{Q}$ is the matrix with eigenvectors as columns and $\mathbf{\Lambda}$ is the diagonal matrix defined by: $\mathbf{\Lambda}=\operatorname{diag}(\lambda_{1},\dots,\lambda_{n})$. Thus we have:

$$\mathbf{A}=\frac{1}{2}\begin{pmatrix}1 & -1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & \sqrt{2}\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4\end{pmatrix}\begin{pmatrix}1 & 1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & \sqrt{2}\end{pmatrix}$$

For the next bit of the question we note that in general we have:

$$\mathbf{A}^{n}=\left(\mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^{-1}\right)^{n}=\mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^{-1}\mathbf{Q}\cdots\mathbf{Q}^{-1}\mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^{-1}=\mathbf{Q}\mathbf{\Lambda}^{n}\mathbf{Q}^{-1},\quad \forall n \in \mathbb{N}$$

Thus we have for your matrix:

\begin{align*}\mathbf{A}^{10}&=\frac{1}{2}\begin{pmatrix}1 & -1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & \sqrt{2}\end{pmatrix}\begin{pmatrix}1^{10} & 0 & 0 \\ 0 & 2^{10} & 0 \\ 0 & 0 & 4^{10}\end{pmatrix}\begin{pmatrix}1 & 1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & \sqrt{2}\end{pmatrix} \\ &= \begin{pmatrix}524800 & 523776 & 0 \\ 523776 & 524800 & 0 \\ 0 & 0 & 1\end{pmatrix}\end{align*}

We note that for the last part we have that for an $n\times n$ matrix, we have that the coefficient of $\lambda^{0}$ is given by $\det(\mathbf{A})$, thus we have:

$$\det(\mathbf{A})=8$$

• Note that some define $\chi_A(\lambda) = \det(\lambda I - A)$. In this case $\det(A) = (-1)^n \chi_A(0)$ instead of $\det(A) = \chi_A(0)$. The former convention results in $\chi_A$ being normalized (The coefficient of $\lambda^n$ will always be $1$). – AlexR Feb 5 '15 at 1:58