Process of finding the eigenvalues of a 3x3 matrix

Question

I'm trying to find the eigenvalues of a 3x3 matrix in order to eventually find an orthogonal matrix $Q$ and diagonal matrix $D$ such that $Q^TAQ = D$, where $A$ is a symmetric matrix, however I'm not too sure on some parts of the process of achieving this:

$A = \begin{pmatrix}3 &2&2 \\ 2&3&-2\\2&-2&3 \end{pmatrix}$

Process:

1. $det(tI - A) = \begin{vmatrix}t-3 &-2&-2\\-2&t-3&2\\-2&2&t-3 \end{vmatrix}$

2. $R1 = R1 + R2$ & $R3 = R3 - R2$

$=> \begin{vmatrix}t-5&t-5&0\\-2&t-3&2\\0&5-t&t-5 \end{vmatrix}$

3. $=> (t-5)^2 \begin{vmatrix}1&1&0\\-2&t-3&2\\0&-1&1 \end{vmatrix}$

4. $R2 = R2 + 2R1$

$=> (t-5)^2 \begin{vmatrix}1&1&0\\0&t-1&2\\0&-1&1 \end{vmatrix}$

5. $=> (t-5)^2 \begin{vmatrix}t-1&2\\-1&1\end{vmatrix}$

6. $=> (t-5)^2 (t+1)$

Thus 5 and -1 are the eigenvalues of $A$.

I don't understand how to get from step 2 to step 3 and from step 4 to step 5, everything else is fine. Thanks for all the help in advance.

Answer 1

In Step 2 to Step 3, $t-5$ is factored out of the first row and the third row. The multiplies the determinant by the same factor, each time the factoring is done.

In Step 4 to Step 5, expansion by minors is done on the first column. This is easy since there is a $1$ in only one position and $0$'s in the other positions in that column. What's more, adding the row and column numbers for the position of that $1$ yields an even number. That means that the determinant of the smaller sub-matrix gotten by removing the first row and first column of the large matrix equals the determinant of the larger matrix.

Answer 2

Your work is correct .

You are calculating a determinant and in the step $2 \rightarrow 3$ is used the fact that a determinant is a multilinear function of the rows ( and of the columns) of the matrix. The same is obtained using the fact that the determinant of a product is the product of the determinants. Explicitly:

$ \det\begin{pmatrix}t-5&t-5&0\\-2&t-3&2\\0&5-t&t-5 \end{pmatrix}= \det\left[\begin{pmatrix}t-5&0&0\\0&1&0\\0&0&t-5 \end{pmatrix}\begin{pmatrix}1&1&0\\-2&t-3&2\\0&-1&1 \end{pmatrix}\right]= $

$=(t-5)^2\det \begin{pmatrix}1&1&0\\-2&t-3&2\\0&-1&1 \end{pmatrix}$

and for $4 \rightarrow 5$ you have developed the determinant with the first column ( see @Rory answer).