Finding matrix A with double eigenvalue -3 and A$\begin{bmatrix}1 \\ 2 \\ -2\end{bmatrix}$ = $\begin{bmatrix}6 \\ 12 \\-12\end{bmatrix}$

Question

A is a symmetric matrix size of 3x3 that has two eigen values of -3 and we have the equation A$\begin{bmatrix}1 \\ 2 \\ -2\end{bmatrix}$ = $\begin{bmatrix}6 \\ 12 \\-12\end{bmatrix}$. The problem requires me to write matrix A in standard form.

I have figured out matrix A has diagonalisation and therefore wrote it as Q=$\begin{bmatrix}x&w&q\\w&y&r\\q&r&z\end{bmatrix}$. Using the given equation I got a set of 3 equations: $$x+2w-2q = 6$$ $$w+2y-2r=12$$ $$q+2r-2z=-12$$

I also found the last eigenvalue was 6 and from there the characteristic polynomial $$(x+3)^2(x-6)$$ I calculated the determinant of matrix of Q and compared the coefficients of the polynomial and got another set of equations: $$w+y+z=0$$ $$w^2 + r^2 + q^2 -xy-xz-yz=27$$ $$2wqr - w^2z - qy + xyz - xr^2 = 54$$

I tried solving the set of these 6 equations but could not figure it out. Is there another way to solve this?

Answer 1

You’re not taking advantage of the properties of symmetric real matrices. You know from the second requirement that the eigenspace $E_6$ of $6$ is spanned by $(1,2,-2)^T$. For a symmetric real matrix, eigenvectors with different eigenvalues are orthogonal, therefore the eigenspace of $-3$ is $E_6^\perp$. So, pick any two linearly-independent vectors that are orthogonal to $(1,2,-2)$ and you have your basis of eigenvalues with which to construct $A$ from its diagonalization. If you also choose these vectors to be orthogonal, that will make inverting the resulting matrix even easier.

Answer 2

Denote $e_3 = [1, 2, -2]/3$, then the last condition amounts to say $e_3$ is a unit eigenvector of $A$, with corresponding eigenvalue $6$. Together with the fact $A$ is symmetric and $-3$ are its two remaining eigenvalues, the spectral theorem asserts that $A$ has form

$$A = -3e_1e_1^T - 3e_2e_2^T + 6e_3e_3^T, \tag{1} $$ where $\{e_1, e_2, e_3\}$ forms an orthonormal group of $\mathbb{R}^3$. That $\{e_1, e_2\}$ is orthogonal to $e_3$ implies $e_1$ and $e_2$ falls in the plane $x_1 + 2x_2 - 2x_3 = 0$. This equation has infinitely many solutions, and it's not hard to pick up two orthonormal ones (just a convenient choice) as: \begin{equation} e_1 = \frac{1}{\sqrt{2}}\begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix}, \quad e_2 = \frac{1}{3\sqrt{2}}\begin{bmatrix}4 \\ -1 \\ 1 \end{bmatrix}. \tag{2} \end{equation}

Plug $e_1, e_2, e_3$ back to $(1)$, it gives \begin{equation} A = \begin{bmatrix} -2 & 2 & -2 \\ 2 & 1 & -4 \\ -2 & -4 & 1 \end{bmatrix}. \end{equation}

Note that $A$ does not depend on the specific choice of $e_1, e_2$ in $(2)$, hence uniquely determined. This is because any other valid eigenvectors $\{e_1', e_2'\}$ must be an orthogonal transformation of $\{e_1, e_2\}$, and $(1)$ is invariant under orthogonal transformation.

Answer 3

The simplest way to solve this problem is as follows:

First, notice that the equation given is an eigenvector equation, saying that the vector $\begin{bmatrix}1 \\ 2 \\ -2\end{bmatrix}$ is an eigenvector with eigenvalue $6$.

Now, observe that if $v$ is an eigenvector $A$ with eigenvalue $\lambda$, then it is an eigenvector of $A+\mu I$ with eigenvalue $\lambda+\mu$, where $I$ is the identity matrix. This is easily seen by direct calculation: $$(A+\mu I)v = Av + \mu I v = \lambda v + \mu v = (\lambda+\mu) v$$

Since all but one eigenvalue of the desired matrix are $-3$, you can therefore write the desired matrix as $A=B-3I$ where $B$ has one eigenvalue of $6 - (-3) = 9$ and the other eigenvalues $0$. In particular, it's a symmetric rank-1 matrix, therefore of the form $v v^T$ with $v$ a multiple of the eigenvector with nonzero eigenvalue.

Now we know that this eigenvector is the given vector $$v=\begin{bmatrix}1 \\ 2 \\ -2\end{bmatrix}$$ and a simple calculation gives $$\begin{bmatrix}1 & 2 & -2\end{bmatrix}\begin{bmatrix}1 \\ 2 \\ -2\end{bmatrix} = 9.$$ Thus we already get the desired eigenvalue $9$ for the rank-1 matrix $$B=\begin{bmatrix}1 \\ 2 \\ -2\end{bmatrix}\begin{bmatrix}1 & 2 & -2\end{bmatrix} = \begin{bmatrix} 1 & 2 & -2\\ 2 & 4 & -4\\ -2 & -4 & 4 \end{bmatrix}$$ Now to get the matrix with the desired properties, we just have to subtract $3I$, which means subtracting $3$ from each diagonal element, thus we arrive at $$A=\begin{bmatrix} -2 & 2 & -2\\ 2 & 1 & -4\\ -2 & -4 & 1 \end{bmatrix}$$