I am solving David Lay's 4th edition 7.1 number 16. So here's the problem. The original matrix: $$\begin{bmatrix}-7 & 24 \\ 24 & 7\end{bmatrix} \ ,$$ with eigenvalues $\pm 25$.
I am having problem calculating eigen values assigned to $-25$. So augmented matrix for $(A+25I)x=0$ is $$\begin{bmatrix}18 & 24 \\ 24 & -18\end{bmatrix} \ .$$
If I use RREF function in my TI-83+ calculator, I get $$\begin{bmatrix}1 & 0 & 0 \\ 1 & 0 & 0\end{bmatrix} \ .$$ So I am only getting zero vector, which is useless because what I need is a nonzero eigenvector. But if I use REF, I get $$\begin{bmatrix} 1 & -0.75 & 0 \\ 0 & 1 & 0\end{bmatrix} \ ,$$ which gives me $\begin{bmatrix}-4 & 3\end{bmatrix}^T$ and $\begin{bmatrix}0 & 0\end{bmatrix}^T$.
My 2 questions are the following:
1) Why can't I get eigenvectors using RREF? I think it had worked for other matrices in the past. does this mean I should only use REF to do this? If so, why?
2) When I use REF, I got zero vector and non zero vector. Will I always get zero vector since zero vector is always in the orthogonal space? (and is this why I got the 0 0 line?)
Best Answer
The eigenvalues and eigenvectors are:
$$\lambda_1 = -25, v_1 = (-4, 3)$$
$$\lambda_2 = 25, v_2 = (3, 4)$$
You wrote the matrix incorrectly and I used that so also made an error! I corrected it below!
For the eigenvalues, we form the system:
$$[A -\lambda I]v = 0$$
So, for $\lambda = -25$, we have:
$$\begin{bmatrix}18 & 24\\24 & 32\end{bmatrix}v = 0$$
The RREF can be arrived at as:
$$\begin{bmatrix}1 & \frac{4}{3}\\0 & 0 \end{bmatrix}v = 0$$
For this eigenvalue, we then have:
$$x = -\frac{4}{3}y$$
We can choose $y$ freely as: $y = 3 \rightarrow x = -4$, so, the eigenvector is $(-4, 3)$ as I showed above.
You need to repeat the steps above for the other eigenvalue and RREF works there (try and and see if you the eigenvector I show above).
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