Find the basis for both eigenvalues

[LosTacos]

Homework Statement

Given matrix A= {[39/25,48/25],[48/25,11/25]} find the basis for both eigenvalues.

Homework Equations

The Attempt at a Solution

I row reduced the matrix and found both eigenvalues. I found λ = -1, and λ = 3. Then, I used diagonalization method [-1I₂ - A 0] and [3I₂ - A 0]. I got a basis of {[1,-3/4]} and {[1,4/3]}. However, I checked these but these were incorrect. I wasn't sure what I did wrong.

 

[HallsofIvy]

Row reduce the matrix? You can't find eigenvalues by row reduction!

If [x, y] is an eigenvector corresponding eigenvalue -1, then
$$\begin{bmatrix}\frac{39}{25} & \frac{48}{25} \\ \frac{49}{25} & \frac{11}{25}\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}-x \\ -y\end{bmatrix}$$

$$\begin{bmatrix}\frac{39}{25}x+ \frac{48}{25}y \\ \frac{48}{25}x+ \frac{11}{25}y\end{bmatrix}= \begin{bmatrix}-x \\ -y \end{bmatrix}$$

So we must have $39x/25+ 48y/25= -x$, which is the same as $$64x+ 48y= 0$$ and $$48x/25+ 11y/25= -y$$ which is the same as $$48x+ 36y= 0$$.

Both of those equations reduce to a single equation of the form ax+ by= 0 so that y= (b/a)x.

If [x, y] is an eigenvector corresponding eigenvalue 3, then
$$\begin{bmatrix}\frac{39}{25} & \frac{48}{25} \\ \frac{49}{25} & \frac{11}{25}\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}3x \\ 3y\end{bmatrix}$$

$$\begin{bmatrix}\frac{39}{25}x+ \frac{48}{25}y \\ \frac{48}{25}x+ \frac{11}{25}y\end{bmatrix}= \begin{bmatrix}-x \\ -y \end{bmatrix}$$

So we must have $39x/25+ 48y/25= 3x$, which is the same as $$-36x+ 48y= 0$$ and $$48x/25+ 11y/25= 3y$$ which is the same as $$48x- 64y= 0$$. Again, both of those equations reduce to a single equation.

 

[LosTacos]

Okay, so given those two solutions, do I row reduce it to determine teh basis?

 

[Mark44]

Okay, so given those two solutions, do I row reduce it to determine teh basis?

No. The first equation in HallsOfIvy's post is 48x + 36y = 0, which is equivalent to 4x + 3y = 0. From that equation you can find the eigenvector that is associated with the eigenvalue λ = -1.

The second equation is 48x - 64y = 0, or 3x - 4y = 0. Use this equation to find the eigenvector that is associated with the eigenvalue λ = 3.

 

[LosTacos]

thank you