Solving for D in a Confusing Matrix Equation | Invertible Matrices NxN

[das1]

Solve the equation below for D. Assume that matrices A and B are both invertible and that all matrices are equivalently shaped as NxN (e.g., 3x3 ...).

Equation: ABDB-1 = I

 

[Fermat1]

Solve the equation below for D. Assume that matrices A and B are both invertible and that all matrices are equivalently shaped as NxN (e.g., 3x3 ...).

Equation: ABDB-1 = I

Your equation doesn't make sense because a matrix minus a scalar has no meaning.

 

[das1]

Whoops i copied and pasted that. The -1 should be an exponant

 

[Fermat1]

Whoops i copied and pasted that. The -1 should be an exponant

Asumming it should read $ABDB^{-1}=I$, post multiply by $B$, then pre multiply by $A^{-1}$ to get $BD=A^{-1}B$ Then pre multiply by $B^{-1}$ to get $D=B^{-1}A^{-1}B$.

 

[das1]

Asumming it should read $ABDB^{-1}=I$, post multiply by $B$, then pre multiply by $A^{-1}$ to get $BD=A^{-1}B$ Then pre multiply by $B^{-1}$ to get $D=B^{-1}A^{-1}B$.

Thanks a lot!

If you have a chance, I'm also really confused with this question:

"Show that λ = 1 an eigenvalue of matrix below. In addition, find one corresponding eigenvector.
The 3x3 matrix A is:
4 -2 3
0 -1 3
-1 2 -2
"

I know I'm supposed to find the determinant of the matrix that's λI₃- A but I'm totally unable to factor it, i can't find any of the eigenvalues. Is there a quicker/easier way to do this because I keep getting tripped up.

 

[Fermat1]

Thanks a lot!

If you have a chance, I'm also really confused with this question:

"Show that λ = 1 an eigenvalue of matrix below. In addition, find one corresponding eigenvector.
The 3x3 matrix A is:
4 -2 3
0 -1 3
-1 2 -2
"

I know I'm supposed to find the determinant of the matrix that's λI₃- A but I'm totally unable to factor it, i can't find any of the eigenvalues. Is there a quicker/easier way to do this because I keep getting tripped up.

You have to show that the matrix has a fixed point $(x,y,z)$. If you write that out, you will be find that you need to find a solution to the system $3x-2y+3z=0$, $3z-2y=0$ and $-x+2y-3z=0$. From the second, we get $y=(3/2)z$. Substituting that into either of the two others gives $x=0$. So (0,3,2) is an eigenvector with eigenvalue 1

 

[Deveno]

Thanks a lot!

If you have a chance, I'm also really confused with this question:

"Show that λ = 1 an eigenvalue of matrix below. In addition, find one corresponding eigenvector.
The 3x3 matrix A is:
4 -2 3
0 -1 3
-1 2 -2
"

I know I'm supposed to find the determinant of the matrix that's λI₃- A but I'm totally unable to factor it, i can't find any of the eigenvalues. Is there a quicker/easier way to do this because I keep getting tripped up.

One typically uses the characteristic polynomial to FIND eigenvalues. If we already KNOW an eigenvalue $\lambda$, then we find eigenvectors by solving:

$(A - \lambda I)v = 0$.

In this case, that means solving:

$\left(\begin{bmatrix}4&2&-3\\0&-1&3\\-1&2&-2 \end{bmatrix} - \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}\right) \begin{bmatrix}v_1\\v_2\\v_3 \end{bmatrix} = \begin{bmatrix}0\\0\\0 \end{bmatrix}$

or:

$\begin{bmatrix}3&2&-3\\0&-2&3\\-1&2&-3 \end{bmatrix} \begin{bmatrix}v_1\\v_2\\v_3 \end{bmatrix} = \begin{bmatrix}0\\0\\0 \end{bmatrix}$

or:

$3v_1+2v_2-3v_3 = 0$

$-2v_2+3v_3 = 0$

$-v_1+2v_2-3v_3 = 0$

As in Fermat's post, it is easy to see that $v_2 = \frac{3}{2}v_3$, so that $v_1 = 0$ (by substitution, or just add equations 1 and 2). Letting $v_3 = 2$ gives one possible eigenvector as (0,3,2).

But, let's see (just for grins) what $\det(I - \lambda A)$ works out to be. The matrix in question is:

$\begin{bmatrix}\lambda-4&-2&3\\0&\lambda+1&-3\\1&-2&\lambda+2 \end{bmatrix}$

which has determinant:

$(\lambda-4)(\lambda+1)(\lambda+2) + 6 - 3(\lambda+1) - 6(\lambda-4)$

$= \lambda^3 - \lambda^2 - 10\lambda - 2 - 3\lambda - 3 - 6\lambda + 24$

$= \lambda^3 - \lambda^2 - 19\lambda + 19$

It is not hard to see $\lambda = 1$ is a root, since this leads to: $1 - 1 - 19 + 19 = 0$.

In fact, $\lambda^3 - \lambda^2 - 19\lambda + 19 = (\lambda - 1)(\lambda^2 - 19)$ so we have two other eigenvalues as well: $\pm \sqrt{19}$

 

[das1]

One typically uses the characteristic polynomial to FIND eigenvalues. If we already KNOW an eigenvalue $\lambda$, then we find eigenvectors by solving:

$(A - \lambda I)v = 0$.

In this case, that means solving:

$\left(\begin{bmatrix}4&2&-3\\0&-1&3\\-1&2&-2 \end{bmatrix} - \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}\right) \begin{bmatrix}v_1\\v_2\\v_3 \end{bmatrix} = \begin{bmatrix}0\\0\\0 \end{bmatrix}$

or:

$\begin{bmatrix}3&2&-3\\0&-2&3\\-1&2&-3 \end{bmatrix} \begin{bmatrix}v_1\\v_2\\v_3 \end{bmatrix} = \begin{bmatrix}0\\0\\0 \end{bmatrix}$

or:

$3v_1+2v_2-3v_3 = 0$

$-2v_2+3v_3 = 0$

$-v_1+2v_2-3v_3 = 0$

As in Fermat's post, it is easy to see that $v_2 = \frac{3}{2}v_3$, so that $v_1 = 0$ (by substitution, or just add equations 1 and 2). Letting $v_3 = 2$ gives one possible eigenvector as (0,3,2).

But, let's see (just for grins) what $\det(I - \lambda A)$ works out to be. The matrix in question is:

$\begin{bmatrix}\lambda-4&-2&3\\0&\lambda+1&-3\\1&-2&\lambda+2 \end{bmatrix}$

which has determinant:

$(\lambda-4)(\lambda+1)(\lambda+2) + 6 - 3(\lambda+1) - 6(\lambda-4)$

$= \lambda^3 - \lambda^2 - 10\lambda - 2 - 3\lambda - 3 - 6\lambda + 24$

$= \lambda^3 - \lambda^2 - 19\lambda + 19$

It is not hard to see $\lambda = 1$ is a root, since this leads to: $1 - 1 - 19 + 19 = 0$.

In fact, $\lambda^3 - \lambda^2 - 19\lambda + 19 = (\lambda - 1)(\lambda^2 - 19)$ so we have two other eigenvalues as well: $\pm \sqrt{19}$

This is great, thank you! I was trying to do way more than I needed for this one. That's some impressive factoring too!