1. ## Diagnolize Matrix

I am preparing to take an exam on Linear Algebra.
I found this question on previous exams.

Here is what I think is needed to be done.
1)First find the eigenvalues, I get:
$\det(kI-A)=0$
$\left| \begin{array}{ccc}k+1&7&-1\\0&k-1&0\\0&15&k+2 \end{array} \right|=0$

$(-1)^{1+1}(k+1)\left| \begin{array}{cc}k-1&0\\15&k+2 \end{array}\right|=(k+1)(k-1)(k+2)=0$
$k=-2,-1,1$

2)There are exactly three eigenvalues, thus there are exactly three linearly independant eigenvectors. Since the eigenvectors are a subspace of the nullspace of $(kI-A)$ which is of dimension three it tells us that there is basis for the eigenspace.

3)To find the basis for the eigenspace I need to substitute the eigenvalues and solve the homogenous system $(kI-A)\bold{x}=\bold{0}$. Thus I need to,
$\left[ \begin{array}{cccc}k+1&7&-1&0\\0&k-1&0&0\\0&15&k+2&0 \end{array} \right]$
Solve it by Gaussian elimination for each eigenvalue $k$.

4)Then the basis I get for the eigenspace in #3 will be the diagnolizable matrix for $A$ by making the vectors as coloum vectors.

Is this approach correct?

I have a second question, how is "Eigenvalue" pronounced.
(At least it is not as bad, Lebesque. I never would have guess it, unless I heard someone pronouce it).

2. Originally Posted by ThePerfectHacker
...

I have a second question, how is "Eigenvalue" pronounced.
...
Hello TPH,

as you may have guessed "eigen" is a German word which means "own". I'll try to transcript it so you can pronounce it:

"ei" pronounce exactly like "I"
"gen" is the unstressed syllable. So substitute in the word "gun" the "u" by an "e" similar to the "e" in "then". That's it.

EB

3. Hello TPH,

in addition to my previous post I have found a better way to help you. I've attached a copy from my English-German dictionary.

EB

4. Your eigenvalues are correct and the method to find the corresponding eigenvectors is also right.
I found the eigenvectors to be {(1,0,1), (1,0,0), (1,1,5)} respectively with the eigenvalues -2, -1, 1.

If you put these eigenvectors, which span the eigenspace, in the columns of a matrix P (which is invertible), then you can write A as PDP^(-1) where D is a diagonal matrix whose elements are the eigenvalues (in the same order as the eigenvectors in P).

5. I got the same results.

One easy way to determine if a matrix is, first of all, diagonalizable is to check that the algebraic multiplicity of the eigenvalues equals the geometric multiplicity, or dimension. That is, it is only diagonalizable if the sum of the dimensions for each distinct eigenspace is equal to n, which occurs if and only if the dimension for the eigenspace of each lambda k is equal to the multicplicty of lambda k.

Recall that a matrix is diagonalizable if it can be written in factored form PDP^(-1), where D is a diagonal matrix and P is an invertible matrix.

In general, to diagonalize an n x n matrix, A:

1.) Find the eigenvalues of A.
2.) Find n linearly independent eigenvectors of A (this step is crucial; if it fails then A cannot be diagonalized)
3.) Construct P from the vectors in the previous step (order is not important)
4.) Construct D from the respective eigenvalues (in this step, it is essential that the order of the eigenvalues matches the order chosen for the columns of P).

And a nice theorem follows that an n x n matrix with n distinct eigenvalues is diagonalizable.

6. I just like to mention, that the matrix,
$PDP^{-1}$
Looks like an inner automorphism to me.
Is this fact of importance in linear algebra?

7. I found it extremely important when studying Lin. Alg.

A = PDP^(-1), with D a diagonal matrix guarantees that you can diagonalize the matrix where the columns of P are n linearly independent vectors of A. The diagonal matrix of D are eigenvalues of A that correspond to the eigenvectors of P. Note that P is any n x n matrix with columns v_1, ..., v_n, and D is any diagonal matrix w/ diagonal entries lambda_1, ..., lambda_n.

Then, AP = A[v_1 v_2 ... v_n] = [Av_1 Av_2 ... Av_n] while

PD = P[[lambda_1 0 ... 0],[0 lambda_2 0 ... 0], ..., [0 0 ... lambda_n]]

= [lambda_1v_1 lambda_2v_2 ... lambda_nv_n]

Suppose that A is diagonalizable and that A = PDP^(-1). We know AP = PD by right-multiplying the above.

[Av_1 Av_2 ... Av_n] = [lambda_1v_1 lambda_2v_2 ... lambda_nv_n]

Thus: Av_1 = lambda_1v_1, Av_2 = lambda_2v_2, ..., Av_n = lambda_nv_n (*)

Since we have that P is invertible, we know that its columns v_1, ..., v_n have to be lin. independent. And since we have that the columns are nonzero, the above (*) shows lambda_1, ..., lambda_n are eigenvalues with v_1, ..., v_n as corresponding eigenvectors. Given any n eigenvecotrs v_1, ..., v_n we can construct the columns of P using the corresponding eigenvalues lambda_1, ..., lambda_n to make D. Recall AP = PD. The eigenvectors are lin. independent, and we know P is invertible (by the invertible matrix theorem), and since AP = PD implies that A = PDP^(-1).