Can I apply elementary row operation then find eigen values of a matrix?

Question

Suppose if a matrix is given as

$$ \begin{bmatrix} 4 & 6\\ 2 & 9 \end{bmatrix}$$

We have to find its eigenvalues and eigenvectors.

Can we first apply elementary row operation . Then find eigenvalues.

Is their any relation on the matrix if it is diagonalized or not.

Answer 1

No, elementary row operations need not preserve eigenvalues and/or eigenvectors.

Examples.

$$\begin{bmatrix}1 & 0 \\0 & 1\end{bmatrix} \overset{R_1 \mapsto R_1 + R_2}{\longrightarrow} \begin{bmatrix}1 & 1 \\0 & 1\end{bmatrix}$$

In this case, the eigenvalue remains the same but the eigenvectors don't. This can be easily observed by the fact that the left matrix is diagonalisable but the right one isn't.

$$\begin{bmatrix}1 & 1 \\0 & 1\end{bmatrix} \overset{R_2 \mapsto R_1 + R_2}{\longrightarrow} \begin{bmatrix}1 & 1 \\1 & 2\end{bmatrix}$$

In this case, the eigenvalues don't remain the same.

Answer 2

As noted, you cannot apply elementary row operations to $A$ and expect the eigenvalues/vectors be preserved. However, you can apply elementary row operations to $|A-\lambda I|=0$ to solve for $\lambda$. In your example, we have:

\begin{align}|A-\lambda I|&=\left|\begin{matrix}4-\lambda&6\\2&9-\lambda\end{matrix}\right|\\&=\frac1{4-\lambda}\left|\begin{matrix}4-\lambda&6\\2(4-\lambda)&\lambda^2-13\lambda+36\end{matrix}\right|\\&=\frac1{4-\lambda}\left|\begin{matrix}4-\lambda&6\\0&\lambda^2-13\lambda+24\end{matrix}\right|\\&=\lambda^2-13\lambda+24\end{align}

if you so desired to do so.

Answer 3

As others have noted you can't apply arbitrary elementary row operations to a matrix and expect the eigenvalues/vectors be preserved. The closest you can do is to apply them to both rows and columns in a specific way as follows.

Consider the matrix $$ T = \begin{bmatrix} 1 & 0\\ \alpha & 1 \end{bmatrix} $$ and its inverse $$ T^{-1} = \begin{bmatrix} 1 & 0\\ -\alpha & 1 \end{bmatrix}. $$

Pre-multiplying a matrix by $T$ is like performing the operation $R_2 \leftarrow R_2 + \alpha R_1$ while post-multiplying by $T^{-1}$ is like performing $C_1 \leftarrow C1 - \alpha C_2$.

Since $A$ and $T A T^{-1}$ are similar, they have the same eigenvalues and eigenvectors. One can apply analogue operations to larger matrices. If several such operations are applied they have to be applied in opposite order among rows and columns, so that $T_1 \dots T_n A T_n^{-1} \dots T_1^{-1}$ is similar to $A$.

Answer 4

First compute the eigenvalues as follows: $$ \begin{align} \det(\lambda I-A)&=\det\left( \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} - \begin{bmatrix} 4 & 6 \\ 2 & 9 \end{bmatrix} \right) = \det \left( \begin{bmatrix} \lambda-4 & -6 \\ -2 & \lambda-9 \end{bmatrix} \right) \\ &= (\lambda-4)(\lambda-9)-(-2)(-6) \\ &= \lambda^2-13\lambda + 24 \end{align} $$ The eigenvalues are then $$ \begin{align} \lambda_{1,2} &= \frac{13 \pm \sqrt{13^2-4\times 24} }{2} \\ &= \frac{13\pm \sqrt{73}}{2} \end{align} $$ Now to compute the eigenvector for $\lambda_1$, we need to solve the following $$ \begin{align} \begin{bmatrix} \lambda_1 - 4 & -6 \\ -2 & \lambda_1 -9 \end{bmatrix} &= \begin{bmatrix} \frac{13 + \sqrt{73} }{2} - 4 & -6 \\ -2 & \frac{13 + \sqrt{73} }{2} -9 \end{bmatrix} \\ &= \begin{bmatrix} \frac{5 + \sqrt{73} }{2} & -6 \\ -2 & \frac{-5 +\sqrt{73} }{2} \end{bmatrix} \end{align} $$ Apply Gaussian elimination (i.e. $R_2=R_2+2\frac{2}{5+\sqrt{73}}R_1)$, we get $$ \begin{bmatrix} \frac{5 + \sqrt{73} }{2} & -6 \\ 0 & 0 \end{bmatrix} $$ So, $x_2$ is a free variable and let $x_2=s$, we $$ \begin{align} \frac{5 + \sqrt{73} }{2} x_1 = 6x_2 \implies x_1 = \frac{12}{5+\sqrt{73}}s \end{align} $$ Finally, $$ x= \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} \frac{12}{5+\sqrt{73}} s \\ s \end{bmatrix} = s \begin{bmatrix} \frac{12}{5+\sqrt{73}} \\ 1 \end{bmatrix} $$ The eigenvector for $\lambda_1$ is $$ v_1 = \begin{bmatrix} \frac{12}{5+\sqrt{73}} \\ 1 \end{bmatrix} $$ Now you can compute the second eigenvector in the same scenario.