How can you solve for the remaining variables when multiplied by zero?

[gabriels-horn]

Homework Statement

The Attempt at a Solution

I haven't tackled anything bigger than a 3x3 matrix. Anyone have any good pointers for reducing this matrix? I'm assuming the quickest way is still going to be the cofactor method?

 

[cronxeh]

Its an upper triangular matrix, so eigenvalues are pretty much on the diagonal here, they are all 2. Taking the determinant yields (2-lambda)^5, setting (2-lambda)^5=0, you get lambda1=lambda2=lambda3=lambda4=lambda5=2.

 

[Landau]

No need to calculate, only to think

The matrix is upper-triangular, what does this immediately tell you about the eigenvalues?

 

[gabriels-horn]

Ok, the discussion about the eigenvalues being the main diagonal values is clear. If the five eigenvalues are = 2, then plugging that value into the eigenvector matrix leaves only the one values in the 5x5 matrix above. Setting the matrix = 0 only leaves the eigenvector = 0; how are there then three linearly independent eigenvectors. What am I overlooking?

 

[ninty]

By definition an eigenvector cannot be the zero vector.
I'm assuming you're using the usual methods I see undergraduates use, (A-xI) = 0
If we represent vectors as (x1,x2,x3,x4,x5) we get x2=0, x5=0
Then solving for x1,x3,x4 we can get 3 linearly independent vectors which form a basis for null(A-xI)

An easy one would be to take 3 vectors from the usual basis.
However the question doesn't really ask for the eigenvectors, just show that they do exist.

 

[Landau]

There are three Jordan Blocks corresponding to eigenvalue 2, so its geometric multiplicity is 3.

 

[HallsofIvy]

From the definition of "eigenvector", if v is an eigenvector with eigenvalue 2, then
$$\begin{bmatrix}2 & 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}\begin{bmatrix}u \\ v \\ x \\ y \\ z\end{bmatrix}= \begin{bmatrix}2u+ v \\ 2v \\ 2x \\ 2y+ z \\ 2z\end{bmatrix}= \begin{bmatrix} 2u \\ 2v \\ 2x \\ 2y \\ 2z \end{bmatrix}$$

which gives the equations 2u+ v= 2u, 2v= 2v, 2x= 2x, 2y+ z= 2y, 2z= 2z. The first equation tells us that v= 0 and the fourth equation that z= 0. But u, x, and y can be any numbers.

 

[gabriels-horn]

By definition an eigenvector cannot be the zero vector.
I'm assuming you're using the usual methods I see undergraduates use, (A-xI) = 0
If we represent vectors as (x1,x2,x3,x4,x5) we get x2=0, x5=0
Then solving for x1,x3,x4 we can get 3 linearly independent vectors which form a basis for null(A-xI)

An easy one would be to take 3 vectors from the usual basis.
However the question doesn't really ask for the eigenvectors, just show that they do exist.

Yes, the (A-xI) = 0 method is the one I used and got x2 = 0 and x5 = 0. How can you solve for x1, x3 and x4 when they are multiplied by zero? Wouldn't that also make them zero?

There are three Jordan Blocks corresponding to eigenvalue 2, so its geometric multiplicity is 3.

This isn't something that we went over in Linear or ODE, can you expand on this thought some more?

 

[gabriels-horn]

From the definition of "eigenvector", if v is an eigenvector with eigenvalue 2, then
$$\begin{bmatrix}2 & 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}\begin{bmatrix}u \\v \\ x \\ y \\ z\end{bmatrix}= \begin{bmatrix}2u+ v \\ 2v \\ 2x \\ 2y+ z \\ 2z\end{bmatrix}= \begin{bmatrix} 2u \\ 2v \\ 2x \\ 2y \\ 2z \end{bmatrix}$$

which gives the equations 2u+ v= 2u, 2v= 2v, 2x= 2x, 2y+ z= 2y, 2z= 2z. The first equation tells us that v= 0 and the fourth equation that z= 0. But u, x, and y can be any numbers.

You just put [/math] on the end instead of [/tex] and I took the zero out of the variable column. That's what I was confused on, the fact that x1, x3 and x4 could be any value. I just assumed that they were also zero since putting 2-2 = 0 for the eigenvector only left the two 1's in the matrix corresponding to x2 = 0 and x5 = 0. Thank you guys for the insight and provoding a good matrix template for future use.

 

[Landau]

Yes, the (A-xI) = 0 method is the one I used and got x2 = 0 and x5 = 0. How can you solve for x1, x3 and x4 when they are multiplied by zero? Wouldn't that also make them zero?

Whatever values they have, they get multiplied by zero resulting in zero, so it doesn't matter what they are. In other words, you are free to choose x1, x3 and x4, which means you can form three linearly independent eigenvectors (e.g. taking x1, x3, x4 equal to 1,0,0, equal to 0,1,0, equal to 0,0,1, gives you three independent eigenvectors.)

This isn't something that we went over in Linear or ODE, can you expand on this thought some more?

Your matrix is already in Jordan Normal Form. The number of Jordan blocks corresponding to a given eigenvalue coincides with the geometric multiplicity of that eigenvalue. In this case there are three such blocks for eigenvalue 2.

 

[gabriels-horn]

Whatever values they have, they get multiplied by zero resulting in zero, so it doesn't matter what they are. In other words, you are free to choose x1, x3 and x4, which means you can form three linearly independent eigenvectors (e.g. taking x1, x3, x4 equal to 1,0,0, equal to 0,1,0, equal to 0,0,1, gives you three independent eigenvectors.)
Your matrix is already in Jordan Normal Form. The number of Jordan blocks corresponding to a given eigenvalue coincides with the geometric multiplicity of that eigenvalue. In this case there are three such blocks for eigenvalue 2.

So the Jordan block is essentially dependent on the 1 values of the superdiagonal, and since x1, x3 and x4 have no entries above them on the superdiagonal, the number of Jordan blocks is equal to 3?

 

[HallsofIvy]

Yes, the (A-xI) = 0 method is the one I used and got x2 = 0 and x5 = 0. How can you solve for x1, x3 and x4 when they are multiplied by zero? Wouldn't that also make them zero?

No, multiplying x1 by 0 does NOT make x1 0, it makes the product 0. All that means is that you don't have that variable in that particular equation.

This isn't something that we went over in Linear or ODE, can you expand on this thought some more?