Thread: Find the matrix

Find the matrix give:
Eigen values are Lamba1 & lambda2 = 1, lambda3 = -2

The eigen spaces are:
E1 = span {[1 1 0],[1 1 1]}, E2 = span {[1 -1 0]}

By normalizing the vectors then computing the matrix A for each composition I obtain:

__|1/2..1/2..0| ___|1/3..1/3..1/3| ___|1/2..-1/2..0|
1*|1/2..1/2..0| +1*|1/3..1/3..1/3| -2*|-1/2..1/2..0|
__|..0....0....0| ___|1/3..1/3..1/3| ___|...0....0....0|


I obtain the matrix:
|-1/6 11/6 1/3|
|11/6 -1/6 1/3|
|1/3 1/3 1/3|


This is the incorrect matrix for the eigenvalues and lambda values given!

I suspect when I formulated the spectral decomposition from the normalized vectors I made and error and thus cascade the incorrect values throughout. Please advise.
Thanks!

The fact that there are 3 independent eigenvectors tells you that this matrix can be diagonalized: If we form the matrix P having the eigenvectors as columns, then $\displaystyle P^{-1}AP= D$ where D is the diagonal matrix having the eigenvalues of A along its main diagonal. Then, multiplying on the left by P and on the right by $\displaystyle P^{-1}$, $\displaystyle A= PDP^{-1}$.

Here, $\displaystyle P= \begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & -1 \\ 0 & 1 & 0\end{bmatrix}$ so we can calculate $\displaystyle P^{-1}= \begin{bmatrix}\frac{1}{2} & \frac{1}{2} & -1 \\ 0 & 0 & 1 \\ \frac{1}{2} & -\frac{1}{2} & 0\end{bmatrix}$.

So $\displaystyle A= \begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & -1 \\ 0 & 1 & 0\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2\end{bmatrix}\begin{bmatrix}\frac{1}{2} & \frac{1}{2} & -1 \\ 0 & 0 & 1 \\ \frac{1}{2} & -\frac{1}{2} & 0\end{bmatrix}$

I do not see a singular matrix here?

Example:
Say the answer was...


3/2 -1/2 0
-1/2 3/2 0
0 0 1

When you calculate theeigen values and vectors for this you would get:

Characteristic polynomial:
x^3 - 4x^2 + 5x - 2
Real eigenvalues:
{1, 1, 2}
Eigenvector of eigenvalue 1:
(1, 1, 0)
Eigenvector of eigenvalue 1:
(0, 0, 1)
Eigenvector of eigenvalue 2:
(-1, 1, 0)

In the above values we see that they do not match the original question asking us.."given lambda 1 &2=1, and lambda 3 =-2)....The vectors are supposed to be 1 1 0, 1 1 1, 1 -1 0.

I believe that I am missing something in your analysis however can tell you it should be a single 3x3 matrix.

Ladies and gentleman:

The original problem remains unsolved and I left out steps showing my work but here is what we need to do. Note: I have an "error" somewhere which does not allow me to obtain the proper symmetrical matrix that provides the given lambda and vector values!!:

We know the given eigenspaces and lambda's - OK.

We need to solve using my method as follows:
1st:
Normalize the vectors:

[1 1 0]

[1 1 1]

[1 -1 0]


Now we must compute the matrix A spectral decomposition:

...| (sqrt2)/2|
1*| (sqrt2)/2| * [(sqrt2)/2 (sqrt2)/2 0 ] +
...| 0 |

...| (sqrt3)/3|
1*| (sqrt3)/3| * [(sqrt3)/3 (sqrt3)/3 (sqrt3)/3] +
....| (sqrt3)/3|


.....| (sqrt2)/2|
-2*|-(sqrt2)/2| * [-(sqrt2)/2 -(sqrt2)/2 0 ]
.....| 0 |



This then provides three matrixes:

|1/2 1/2 0| ...|1/3 1/3 1/3| ......| 1/2 -1/2 0|
|1/2 1/2 0| + |1/3 1/3 1/3| -2* | -1/2 -1/2 0|
| 0 0 0.....| ...|1/3 1/3 1/3| .....| 0 0 0|

I got the matrix answer:

|-1/6 11/6 1/3|
|11/6 -1/6 1/3|
| 1/3 1/3 1/3|

When I run this through I can see this does NOT get the lambda and vectors which means the solution matrix I provided is incorrect!

Who can solve this?
Thanks for all the effort folks.

You didn't even check to see if the given vectors are eigenvectors of the matrix I gave for A! The fact that you come up with other eigenvectors doesn't mean the given vectors are not eigenvectors. Any eigenvalue has an infinite number of corresponding eigenvectors.

I gave
$\displaystyle
A= \begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & -1 \\ 0 & 1 & 0\end{bmatrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2\end{bmatrix}\begin{bmatrix}\frac{1}{2} & \frac{1}{2} & -1 \\ 0 & 0 & 1 \\ \frac{1}{2} & -\frac{1}{2} & 0\end{bmatrix}
$
so that $\displaystyle A= \begin{bmatrix}-\frac{1}{2} & \frac{3}{2} & 0 \\ -\frac{3}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & 1\end{bmatrix}$

Now $\displaystyle \begin{bmatrix}-\frac{1}{2} & \frac{3}{2} & 0 \\ -\frac{3}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}= \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$ so $\displaystyle \begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix}$ is an eigenvector with eigenvalue 1.
$\displaystyle \begin{bmatrix}-\frac{1}{2} & \frac{3}{2} & 0 \\ -\frac{3}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}= \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$ so $\displaystyle \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$ is also an eigenvector with eigenvalue 1.

Finally, $\displaystyle \begin{bmatrix}-\frac{1}{2} & \frac{3}{2} & 0 \\ -\frac{3}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}= \begin{bmatrix}-2 \\ 2 \\ 0\end{bmatrix}$ so $\displaystyle \begin{bmatrix} 1 \\ -1 \\ 0\end{bmatrix}$ is an eigenvector with eigenvalue -2.

What more do you want?

You do understand, don't you, that any linear combination of the first two eigevectors will still be an eigenvector with eigenvalue 1, and that any multiple of the third eigenvector will be an eigenvector with eigenvalue 2. Your (1, 1, 2)= (-1)(1, 1, 0)+ 2(1, 1, 1) while (-1, 1, 0) is just -1 times (1, -1, 0).

Here is what the problem is with your answer:

You need to find a projection matrix for each eigen vector "u" (normalized) to be a unit vector...."and" the unit eigenvectors for a "symmetric" matrix A will form an orthogonal basis.

This can involve requirement to use Gram Schmidt's process.

Anyway,
What ever "you" have posted is incorrect. Clearly, the matrix you came up with does not equate to the span shown?? Regardless if the Lambda eigen values are indeed correct.

Here is the form we would nee to run with:

Pv(u) = (u.v)*v
and:
A=L1*P(V1) + L2*P(V2) + L3*P(V3).

Nice try though...
I should not need to state in my question what process to use by the way. The question on the first post is the way it comes but you need to "go back" and confirm that your answer "does" actually match the given initial conditions.

Bottom line:
A=P*D*P^-1 does "not" work in this case.

The question asked intially was

"Find the matrix given:
Eigen values are Lamba1 & lambda2 = 1, lambda3 = -2

The eigen spaces are:
E1 = span {[1 1 0],[1 1 1]}, E2 = span {[1 -1 0]}"

The matrix $\displaystyle
A= \begin{bmatrix}\frac{3}{2} & -\frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{3}{2} & 0 \\ 0 & 0 & 1\end{bmatrix}
$

certainly does satisfy that. No, you should not need to "state in my question what process to use"- not if you are willing to accept any process that gives the correct answer. But if you require that a specific process be used, yes, you do need to tell us that- we can't read your mind!

"The question on the first post is the way it comes but you need to "go back" and confirm that your answer "does" actually match the given initial conditions."
In my last post, I did exactly that.

I do not understand your objection.

Sorry, but I'm closing this thread before things get heated.

@orendacl: Next time try to have a bit more gratitude towards someone who has given a considerable amount of their valuable time and expertise to help you.

Edit: Opened by request from HallsofIvy. When I have time I'll take a closer look at the question and replies.

mr fantastic pointed out to me that I mistakenly used eigenvalue "2" instead of "-2". I have gone back and edited my posts to correct that.

Gentlemen, many thanks for sorting this out. I must apologize for my using an opinionated response(s). It was not my intention to upset and next time I will ensure frustration with a problem does not spill over to those of you who help solve problems!

Originally Posted by orendacl

Gentlemen, many thanks for sorting this out. I must apologize for my using an opinionated response(s). It was not my intention to upset and next time I will ensure frustration with a problem does not spill over to those of you who help solve problems!

No problem. 1) I made an copying mistake. 2) You apparently were taught a specific method for solving a problem like this. I used a different (in my not-so-humble opinion, better) method.

Of course, we can't know what method you need unless you tell us. That's one very good reason (among many) for showing as much as you can of how YOU attempt a problem