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Homework Help: Diagonalizability Quest

  1. Feb 22, 2012 #1
    Show that matrix A = \begin{bmatrix} a&b \\0&a \end{bmatrix} in M2 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?
  2. jcsd
  3. Feb 22, 2012 #2


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    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?
  4. Feb 22, 2012 #3
    For your first question, I am assuming you obtained the characteristic polynomial of [itex](\lambda - a)^2 = 0[/itex].

    So we know we have a repeated eigenvalue.

    0 & b\\
    0 & 0

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

    What does this matrix tell us?

    For one, [itex]x_1[/itex] is a free variable. If b isn't a free a variable, how many eigenvectors would we have?
  5. Feb 22, 2012 #4
    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.
  6. Feb 22, 2012 #5
    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?
  7. Feb 22, 2012 #6


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    ok so you get [itex] \lambda = a [/itex] 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
    \begin{bmatrix} a&b \\0&a \end{bmatrix}\begin{bmatrix} x \\y \end{bmatrix} = a \begin{bmatrix} x \\ y\end{bmatrix}

    which gives
    \begin{bmatrix} ax+by = ax\\ay =ay \end{bmatrix}

    so what constraints does [itex] b \neq 0 [/itex] give?
    Last edited: Feb 22, 2012
  8. Feb 22, 2012 #7
    I don't think I follow... what is the (1,0)^T? What's the raised to the T part?
  9. Feb 23, 2012 #8


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    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).
  10. Feb 24, 2012 #9

    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
  11. Feb 25, 2012 #10


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