Given matrix $A$, find a matrix $S$ such that result of $S^{-1} \circ A \circ S$ is a diagonal matrix

Question

Given is matrix $A=\begin{pmatrix} 3 & 0 & 7\\ 0 & 1 & 0\\ 7 & 0 & 3 \end{pmatrix}$, find a matrix $S$ such that $D=S^{-1} \circ A \circ S$ where $D$ is a diagonal matrix.

So I'm not sure how this is supposed to work, I tried to create some linear system but soon realized there are too many unknowns! Then I thought about the zero matrix, if it's actually a diagonal matrix. This should be the case because a diagonal matrix is a matrix where all non-diagonal entries are zero and this is satisfied in a zero matrix.

But the bad thing is when we want the inverse of that zero matrix, we have a problem because division by zero doesn't work : /

So I can't use that matrix. But how else do you solve this problem?

Answer 1

Find first the eigenvalues of $A$, these are the roots of the characteristic polynomial of $A$, $$ (x-10)(x-1)(x+4)\ . $$ So the eigenvalues are $10$, $1$, $-4$. We subtract them from the diagonal and search for eigenvectors. Here are my choices: $$ \begin{aligned} \begin{bmatrix} 3-10 & 0 & 7\\ 0& 1-10 & 0\\ 7 & 0 & 3-10 \end{bmatrix} \begin{bmatrix} 1\\0\\1 \end{bmatrix} &= \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \ ,\\ \begin{bmatrix} 3-1 & 0 & 7\\ 0& 1-1 & 0\\ 7 & 0 & 3-1 \end{bmatrix} \begin{bmatrix} 0\\1\\0 \end{bmatrix} &= \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \ , \\ \begin{bmatrix} 3-(-4) & 0 & 7\\ 0& 1-(-4) & 0\\ 7 & 0 & 3-(-4) \end{bmatrix} \begin{bmatrix} 1\\ 0\\ -1 \end{bmatrix} &= \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \ ,\\ \end{aligned} $$ Equivalently, $$ \begin{aligned} \begin{bmatrix} 3 & 0 & 7\\ 0& 1 & 0\\ 7 & 0 & 3 \end{bmatrix} \begin{bmatrix} 1\\0\\1 \end{bmatrix} &= \begin{bmatrix} 1\\0\\1 \end{bmatrix} \cdot[10] \ ,\\ \begin{bmatrix} 3 & 0 & 7\\ 0& 1 & 0\\ 7 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0\\1\\0 \end{bmatrix} &= \begin{bmatrix} 0\\1\\0 \end{bmatrix} \cdot [1] \ , \\ \begin{bmatrix} 3 & 0 & 7\\ 0& 1 & 0\\ 7 & 0 & 3 \end{bmatrix} \begin{bmatrix} 1\\ 0\\ -1 \end{bmatrix} &= \begin{bmatrix} 1\\ 0\\ -1 \end{bmatrix} [-4] \ ,\\ \end{aligned} $$ equalities of products of matrices of the shape $(3\times 3)\cdot(3\times 1)$, and respectively $(3\times 1)\cdot(1\times 1)$.

Now consider the matrix built from the eigenvectors, taken as columns. We have the block matrix computation, subsumming the above three equalities: $$ \underbrace{ \begin{bmatrix} 3 & 0 & 7\\ 0& 1 & 0\\ 7 & 0 & 3 \end{bmatrix}}_A \underbrace{ \begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 1& 0& -1 \end{bmatrix}}_S = \underbrace{ \begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 1& 0& -1 \end{bmatrix}}_S \underbrace{ \begin{bmatrix} 10 & & \\ &1 & \\ & & -4 \end{bmatrix}}_D \ . $$ We have $AS=SD$, so the above choices answer the request.

Answer 2

HINT

1. Find the eigenvalues by $|A-\lambda I|=0$

2. For each eigenvalue find the corresponding eigenvector(s) by $(A-\lambda_i I)v_i=0$

3. Consider the matrix $S=[v_1\, v_2\,v_3]$ and calculate $S^{-1}$

4. Then $D=S^{-1}AS$

Answer 3

Let $S$ be the matrix whose columns are eigenvectors of $A$

Then you will have $$ AS=SD$$ where $D$ is the diagonal matrix with eigenvalues on the main diagonal.

Thus $$ S^{-1}AS=D$$ is diagonal.