Different eigenvalues using two different methods

Question

I have to find the Eigen values of the following matrix: $$ \begin{bmatrix} 2 & -1 & 0 & 0\\ 0 & 3 & 0 & 0\\ 0 & 0 & -2 & 0\\ 0 & 0 & -1 &4 \end{bmatrix} $$

I used row operations : R4->R3-2R4

and I got : $$ \begin{bmatrix} 2 &-1 & 0 & 0\\ 0 & 3 & 0 & 0\\ 0 & 0 & -2 & 0\\ 0 & 0 & 0 & -8 \end{bmatrix} $$ Now, for a triangular matrix, Eigen values are the diagonal elements. Hence, E.V = {2,3,-2,-8}

But we also now that,

$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$SUM of Eigen values = Trace(Matrix)

$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$PRODUCT of Eigen values = Det(Matrix).

But by that logic the above mentioned Eigen values are not satisfied. Am i doing something wrong?

Do you want the eigenvalues of the original matrix or the transformed matrix? — peterwhy, Dec 28 '23 at 15:55
@AntonVrdoljak It is upper triangular, as all entries below the diagonal are $0$. — Divide1918, Dec 28 '23 at 15:58
Now you have a counterexample that they are not always the same. Even if you perform row operations on $I$ to obtain $2I$, one matrix satisfies $Iv=1v$ while the other satisfies $(2I)v=2v$, for all vectors $v$. — peterwhy, Dec 28 '23 at 16:03
Row operations can transform any non singular matrix to the identity matrix, so row operations don't preserve eigenvalues. Note $R_4 \to (-1/2)R_3+R_4$ changes R_4 to $0,0,0,4$. — TurlocTheRed, Dec 28 '23 at 17:26

Nothing special · Answer 1 · 2023-12-28T19:09:05.030

As pointed in the comments, row operations, in general, do not preserve eigenvalues.

The operation $R_4\to R_3- 2R_4 $ which you applied, does not even preserve the determinant... The determinant after this row operation is scaled by a factor of $-2$.

$$A=\begin{bmatrix} \color{red}2 & \color{red}{-1} & 0 & 0\\ \color{red} 0 & \color{red} 3 & 0 & 0\\ 0 & 0 & \color{blue}{-2} & \color{blue}{0}\\ 0 & 0 & \color{blue}{-1} &\color{blue}{4} \end{bmatrix}$$

You can easily find the eigenvalues of this matrix by observing that it is block diagonal matrix (consisting of two $2\times 2$ blocks along the diagonal).

The red block is upper triangular and has eigenvalues $2$ and $3$.

The blue block is lower triangular and has eigenvalues $-2$ and $4$.

Thus, the eigenvalues of $A$ are $2$, $3$, $-2$, $4$. See this why.

Different eigenvalues using two different methods

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