What is the Next Step for Finding Linearly Independent Eigenvectors?

[trap101]

Show that matrix A = \begin{bmatrix} a&b \\0&a \end{bmatrix} in M_{2 x 2}(R) is diagonalizable iff b = 0

Attempt: Now I tried to solve for the eigenvalues and eigenvectors, which gave me a matrix of this form: \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

but this matrix won't provide me with any information to show diagonalizability. What am I missing?


Quest 2: Find the necessary and sufficient conditions on the real numbers a,b,c for the matrix:
\begin{bmatrix} 1 & a & b\\ 0 & 1 & C \\ 0 & 0 & 2 \end{bmatrix} to be diagonalizable.

Attempt: Now for this one I also solved for the eigenvlues which were: λ₁ = 1, λ₂ = 1, λ₃ = 2

So the problematic eigenvalues will be the one of multiplicity 2, i.e λ = 1.

So this means I'd have to obtain two linearly independent eigenvectors for λ = 1.

I tried solving and got to this matrix: \begin{bmatrix} 0 & a & b \\ 0&0&c \\ 0&0 & 1 \end{bmatrix}

But I won't be able to find two linearly independent eigenvectors from setting any of the variables equal to anything...I don't think. What's the next step?

 

[lanedance]

Show that matrix A = \begin{bmatrix} a&b \\0&a \end{bmatrix} in M_{2 x 2}(R) is diagonalizable iff b = 0

Attempt: Now I tried to solve for the eigenvalues and eigenvectors, which gave me a matrix of this form: \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

but this matrix won't provide me with any information to show diagonalizability. What am I missing?

b=0 ->
well this one is as the matrix is already diagonal

A is diagonalisable ->
well what do you get for the chracteristic equation

hint: i get a repeated eigenvalue - this means the eigenspace must span R^2 for the matrix to be diagonalisable - can you show this is only possible when b=0?

 

[fauboca]

For your first question, I am assuming you obtained the characteristic polynomial of (\lambda - a)^2 = 0.

So we know we have a repeated eigenvalue.

<br /> \begin{bmatrix}<br /> 0 &amp; b\\<br /> 0 &amp; 0<br /> \end{bmatrix}<br />

Nothing extreme has been done. Simply following the protocol to diagonalize.

What does this matrix tell us?

For one, x_1 is a free variable. If b isn't a free a variable, how many eigenvectors would we have?

 

[trap101]

If b isn't a free variable, then I only have one eigenvector, but my multiplicity is 2 which implies that the matirx isn't diagonalizable. but if b = 0 as in the question then i obtain a diagonal matrix. As for showing it, the only way I could show it is if I obtain the 0 matrix that I talked about below.

 

[fauboca]

If b isn't a free variable, then I only have one eigenvector, but my multiplicity is 2 which implies that the matirx isn't diagonalizable. but if b = 0 as in the question then i obtain a diagonal matrix. As for showing it, the only way I could show it is if I obtain the 0 matrix that I talked about below.

The only value that allows x_2 to be a free variable is 0. With out it, you said you only have one eigenvector. You need n eigenvectors for a nxn matrix. If you only have 1 eigenvector, you have a problem. How does knowing this not help you?

 

[lanedance]

ok so you get \lambda = a with algebraic multiplicity 2

to be diagonalisable you have to have and eigenspace that spans R^2, and as such there exists two linearly independent eigenvectors

so you can represent your eigenvectors as (1,0)^T and (1,0)^T, and x(1,0)^T +y(1,0)^T =(x,y)^T is still an eignevector

then, any eigenvector (x,y)^T must then satisfy
<br /> \begin{bmatrix} a&amp;b \\0&amp;a \end{bmatrix}\begin{bmatrix} x \\y \end{bmatrix} = a \begin{bmatrix} x \\ y\end{bmatrix}<br />

which gives
<br /> \begin{bmatrix} ax+by = ax\\ay =ay \end{bmatrix}<br />

so what constraints does b \neq 0 give?

 

[trap101]

ok so you get \lambda = a with algebraic multiplicity 2

to be diagonalisable you have to have and eigenspace that spans R^2, and as such there exists two linearly independent eigenvectors

so you can represent your eigenvectors as (1,0)^T and (1,0)^T, and x(1,0)^T +y(1,0)^T =(x,y)^T is still an eignevector

then, any eigenvector (x,y)^T must then satisfy
<br /> \begin{bmatrix} a&amp;b \\0&amp;a \end{bmatrix}\begin{bmatrix} x \\y \end{bmatrix} = a \begin{bmatrix} x \\ y\end{bmatrix}<br />

which gives
<br /> \begin{bmatrix} ax+by = ax\\ay =ay \end{bmatrix}<br />

so what constraints does b \neq 0 give?

I don't think I follow... what is the (1,0)^T? What's the raised to the T part?

 

[Deveno]

it means "transpose" and just means we are talking about a column vector, rather than a row vector (since we're dealing with matrices, where rows and columns are different things).

 

[trap101]

ok so you get \lambda = a with algebraic multiplicity 2

to be diagonalisable you have to have and eigenspace that spans R^2, and as such there exists two linearly independent eigenvectors

so you can represent your eigenvectors as (1,0)^T and (1,0)^T, and x(1,0)^T +y(1,0)^T =(x,y)^T is still an eignevector

then, any eigenvector (x,y)^T must then satisfy
<br /> \begin{bmatrix} a&amp;b \\0&amp;a \end{bmatrix}\begin{bmatrix} x \\y \end{bmatrix} = a \begin{bmatrix} x \\ y\end{bmatrix}<br />

which gives
<br /> \begin{bmatrix} ax+by = ax\\ay =ay \end{bmatrix}<br />

so what constraints does b \neq 0 give?

Well simplifying I obtain y = 0, does this mean y has to equal zero in order for the expression to be satisfied? but if that happens then ay = ay becomes 0 = 0

 

[lanedance]

ok so if y = 0, what is the dimension of the eigenspace (how many linearly independent eigenvectors do you have?)