Linear algebra

TheSetMiner

For which values of α do the vectors (1, 1, 1, −1), (1, 2, 2, −1), (3, 6, α, −3),
(1, 1, 1, 1) form a basis for R4?

Am I correct in thinking this problem involves forming a matrix of these vectors, reducing it to REF and solving the reduced equations for α?

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Give an example, with proof, of a matrix with real eigenvalues that is not
diagonalizable.

I know that shear matrices are a good example of this but I don't fully understand why or how to prove this?

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As you can see I am struggling with some of my linear algebra homework and would greatly appreciate any help that can be offered to me!

TheBody

TheSetMiner wrote: »

For which values of α do the vectors (1, 1, 1, −1), (1, 2, 2, −1), (3, 6, α, −3),
(1, 1, 1, 1) form a basis for R4?

Am I correct in thinking this problem involves forming a matrix of these vectors, reducing it to REF and solving the reduced equations for α?

---

Give an example, with proof, of a matrix with real eigenvalues that is not
diagonalizable.

I know that shear matrices are a good example of this but I don't fully understand why or how to prove this?

---

As you can see I am struggling with some of my linear algebra homework and would greatly appreciate any help that can be offered to me!

For the first one, your plan sounds good. Give it a go and see what happens.

For the second one, have a read of this:

http://en.wikipedia.org/wiki/Diagonalizable_matrix

Section "4.3 How to Diagonalize a Matrix" should give you a big hint.

TheSetMiner

TheBody wrote: »

For the first one, your plan sounds good. Give it a go and see what happens.

For the second one, have a read of this:

http://en.wikipedia.org/wiki/Diagonalizable_matrix

Section "4.3 How to Diagonalize a Matrix" should give you a big hint.

OK, for the first one I reduced it to REF and then from reading something online, I found the determinant by multiplying the diagonal entries and got this to equal a/3 - 2. for this to be a basis it should be invertible? and so this must be nonzero. hence a/3 not equal to 2 => a not equal to 6

TheBody

TheSetMiner wrote: »

OK, for the first one I reduced it to REF and then from reading something online, I found the determinant by multiplying the diagonal entries and got this to equal a/3 - 2. for this to be a basis it should be invertible? and so this must be nonzero. hence a/3 not equal to 2 => a not equal to 6

Did you remember that Gaussian elimination has an effect on the det? You would need to keep account of these changes as you converted the matrix to Ref.

If you did, then you should be ok.

TheSetMiner

TheBody wrote: »

For the first one, your plan sounds good. Give it a go and see what happens.

For the second one, have a read of this:

http://en.wikipedia.org/wiki/Diagonalizable_matrix

Section "4.3 How to Diagonalize a Matrix" should give you a big hint.

For the diagnolization, I guess I need to find the suitable eigenvalues and eigenvectors and then set up a (P^-1)AP combination to show it can't equal the diagonal matrix with the eigenvalues as diagonal entries. I still am unsure how I pick my suitable eigenvalues to begin with?

I notice it says there should be exactly n distinct eigenvalues in a nxn matrix for it to be diagnizable? does this mean that say a matrix with a repeated eigenvalue eg. e-values ={1, 1, 2} would be non diagnizable?

TheSetMiner

TheBody wrote: »

Did you remember that Gaussian elimination has an effect on the det? You would need to keep account of these changes as you converted the matrix to Ref.

If you did, then you should be ok.

I used a technique I have seen online where one reduces the matrix to REF and then multiplies the diagonal entries to find the det.? checking the original matrix with a = 6 on an online calculator gives det = 0 which I think proves it correct

TheBody

TheSetMiner wrote: »

For the diagnolization, I guess I need to find the suitable eigenvalues and eigenvectors and then set up a (P^-1)AP combination to show it can't equal the diagonal matrix with the eigenvalues as diagonal entries. I still am unsure how I pick my suitable eigenvalues to begin with?

I notice it says there should be exactly n distinct eigenvalues in a nxn matrix for it to be diagnizable? does this mean that say a matrix with a repeated eigenvalue eg. e-values ={1, 1, 2} would be non diagnizable?

Bingo!!

TheBody

TheSetMiner wrote: »

I used a technique I have seen online where one reduces the matrix to REF and then multiplies the diagonal entries to find the det.? checking the original matrix with a = 6 on an online calculator gives det = 0 which I think proves it correct

It is possible to fluke your answer!

For example, if you divide a row by 2, this will halve the det.

Swapping two rows will multiply the det by -1.

Adding or subtracting multiples of rows have no effect on the det.

I't may be the case that you, for example, swapped rows twice and thus cancelled out the -1 business. It would still need to be accounted for in a solution.

Michael Collins

TheSetMiner wrote: »

...does this mean that say a matrix with a repeated eigenvalue eg. e-values ={1, 1, 2} would be non diagnizable?

This is not necessarily true! It's possible for the matrix to be diagonalisable even in this case.

It won't be diagonalisable provided the dimension of the eigenspace i.e. [latex] A- I\lambda [/latex] is less than the number of times that eigenvalue appears.

For example

[latex] \left[ \matrix{5 & 8 & 16 \cr 4 & 1 & 8 \cr -4 & -4 & -11} \right] [/latex]

has the eigenvalues [latex] \lambda = \{-3,-3, 1\} [/latex]

The eigenvector corresponding to [latex] \lambda = -1 [/latex] is

[latex] v_1 = \left[ \matrix{-2 \cr -1 \cr 1}\right] [/latex]

which corresponds to a one-dimensional space.

But for [latex] \lambda = -3 [/latex], we get

[latex] A- I\lambda = A+ 3I = \left[ \matrix{8 & 8 & 16 \cr 4 & 4 & 8 \cr -4 & -4 & -8} \right] [/latex],

clearly all rows are linearly dependent, so we need to introduce two parameters

[latex] x_3 = t, x_2 = s [/latex].

This gives a two-dimensional space:

[latex] \left[ \matrix{-2 \cr 0 \cr 1}\right]t + \left[ \matrix{-1 \cr 1 \cr 0}\right] s[/latex],

and from this we can get two linearly independent eigenvectors (from a single eigenvalue!).

So, in summary, you need to make sure

1) you have a double root in the characteristic equation, i.e. a repeated eigenvalue
2) you don't get a two-dimensional space from that single eigenvalue, i.e. you need to make sure that [latex] A- I\lambda [/latex] is of rank 2 (not 1, like the example above).