Null space and eigenspace of diagonal matrix

newgrad
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Homework Statement



I am working on a problem where I made a matrix representation of a linear transformation and I am asked what is the eigenspace for a particular eigenvalue.



Homework Equations





The Attempt at a Solution


The problem for me is, I came out with a diagonal matrix. what is the null space of a diagonal matrix? Is it just the 0 vector? If so, can I have an eigenspace?
 
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newgrad said:

Homework Statement



I am working on a problem where I made a matrix representation of a linear transformation and I am asked what is the eigenspace for a particular eigenvalue.



Homework Equations





The Attempt at a Solution


The problem for me is, I came out with a diagonal matrix. what is the null space of a diagonal matrix? Is it just the 0 vector? If so, can I have an eigenspace?

Suppose your diagonal matrix is 3x3. Think about what happens when you multiply your matrix by the vectors [1,0,0], [0,1,0] and [0,0,1]. Aren't those all eigenvectors? What are their eigenvalues in terms of the diagonal matrix entries? A null space is just the set of vectors that have zero eigenvalue, right?
 
The eigenvalues of a diagonal matrix are on the diagonal.The eigenspace of an eigenvalue a is the nullspace of A-aI,that is,the solutions of (A-aI)x=0
 
What do you mean by "came out with" a diagonal matrix? IF the matrix you are dealing with is diagonal, then its eigenvalues are the numbers on the diagonal as Hedipaldi said. An if is the diagonal number in the ith row, its eigenspace is spanned by the ith column. (If the same eigenvalue appears k times on the diagonal, the eigenspace is the space spanned by those k columns.)

But is "came out with" a diagonal matrix means you derived a diagonal matrix somehow from the given matrix we would have to know how it was derived.
 
thanks. I think I am seeing the light. So when I subtract off the eigenvalue times I, then i get a 0 in that spot, so the eigenvector can have a 1, or anything, in that corresponding spot. so the eigenspace would be all the vectors with the basis of all O's and a 1 in a column, and the one is where that particular eigenvalue was. did that make sense?
 
newgrad said:
thanks. I think I am seeing the light. So when I subtract off the eigenvalue times I, then i get a 0 in that spot, so the eigenvector can have a 1, or anything, in that corresponding spot. so the eigenspace would be all the vectors with the basis of all O's and a 1 in a column, and the one is where that particular eigenvalue was. did that make sense?

Not really. That may not be the clearest explanation. But if it's making sense to you, that's what counts.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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