Consider the matrix: $$M=\begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}$$ (i) Determine the two eigenvalues, and the normalised eigenvectors

(ii)Establish whether the eigenvectors are orthogonal to each other, or not

(iii)Write the normalised eigenvectors into the columns of a matrix D, and show that the inverse matrix $D^{-1}$ us given by its transpose $D^{T}$. Hint: compute the matrix product $D .D^T$

(iv) Show that the matrix: $$M=D.M.D^{-1}$$ is a diagonal matrix with the eigenvalues on the diagonal

(v)Consider the matrix

$$N=\frac{1}{2} \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}$$ Show that N is the inverse matrix to M, and that the eigenvectors of M are also eigenvectors of N, and read off the eigenvalues.

My Attempt
$$M\vec{e}=\mu \vec{e}$$ where $\vec{e}$ is an eigenvector and $\mu$ is an eigenvalue. $$A\vec{e}-\mu I \vec{e}=0,$$ $$(A-\mu I) \vec{e}=0.$$ If $\vec{e}$ is non-zero then this means that the columns of $A-\mu I$ are linearly dependent. Hence $$det(A-\mu I)=0,$$ $$det(A-\mu I)= \begin{vmatrix} -\mu & 2 \\ 2 & -\mu \end{vmatrix}$$ From this, we see that: $$\mu=\pm 2,$$ these are the eigen values.

To find the associated eigenvectors, I re-substitute these back into: $$(A-\mu I) \vec{e}=0.$$ For $\mu$=2, we get: $$ \begin{bmatrix} -2 & 2 \\ 2 & -2 \end{bmatrix} . \begin{bmatrix} x_1\\ y_1 \end{bmatrix}=0$$ For $\mu$=-2, we get: $$ \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} .\begin{bmatrix} x_2 \\ y_2 \end{bmatrix}=0$$ For $\mu=2$, this is the same as writing: $$-2x_1+2y_1=0,$$ $$2x_1-2y_1=0.$$ here, we see that an eigenvector satisfying both of these are: $$\begin{bmatrix} x_1 \\ y_1 \end{bmatrix}=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$$ For $\mu=-2$, this is the same as writing: $$2x_2+2y_2=0,$$ $$2x_2+2y_2=0.$$ here, we see that an eigenvector satisfying both of these are: $$\begin{bmatrix} x_2 \\ y_2 \end{bmatrix}=\begin{bmatrix} 1 \\ -1 \end{bmatrix}$$ and to normalise them I can just put $\frac{1}{\sqrt{2}}$ in front of both of them.

Problem #1

Although I think I am right with these eigenvectors it does confuse me that I could flip the signs round and still have an eigenvector that satisfies my equations, for example in the case where $\mu=-2$ surely the eigenvector could just as easily be: $$\begin{bmatrix} x_2 \\ y_2 \end{bmatrix}=\begin{bmatrix} -1 \\ 1 \end{bmatrix}$$ I understand that when I normalise the vector, I am essentially making it a unit vector... but surely flipping the signs as I have just now would flip the direction of the unit vector, essentially making it a different one. I am thinking that it is my decision as to what I do with this, as I would be choosing them for the base of a new co-ordinate system, but I was hoping someone could verify whether or not I am correct in thinking this.

Part(ii) I don't have a problem with (just use dot product)

Part(iii) I have generated the matrix: $$D=\begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$$ By simply putting the eigenvectors in the columns of the matrix. I have computed the inverse of the matrix, by the following method: $$\frac{1}{det(D)}*Adj{D}$$ $$D^{-1}=-1. \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}$$ OR $$ D^{-1}=\begin{bmatrix} -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$$

Howver, If we consider the transpose of the matrix D, we get that: $$D^T=\begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$$ Problem #2

I have clearly gone wrong, as by my calculation $D^{-1}$ is not equal to $D^T$ I was wondering if anyone could spot my mistake, as I have repeated it several times I can not see. Also, I am unsure why the question has told me to compute $D^{-1}*D^{T}$, perhaps this is some identity that I am unaware of?

Problem 3

For the other two questions I was just wondering if someone could let me know what would be sufficient to satisfy the 'show that', can i simply computw the matrix multiplication of $D.M.D^{-1}$? and the same for part (v). Basically I am asking if there are any Identities that would allow me to show it without computing it all.

Sorry for the long post, but I would really appreciate if someone could help me with this. Thanks :)


Problem #1: when you change the sign, what you get is a vector that is collinear to the one you had. Therefore there is no problem at all, you know that eigenspaces are linear subspaces.

Problem #2: there is a sign issue when you compute $D^{-1}$, at the very first step. Adj means the transpose of the comatrix.

Problem #3: (iv) you can avoid computing the product by using the fact that the columns of $D$ are eigenvectors. Hint: the first column of a matrix $M$ is in fact $M\cdot \left(\matrix{1\\0} \right)$.

  • $\begingroup$ Thanks, I have re-attempted to calculate the inverse matrix of D, I have gotten: $D=\begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$. I think that I am getting the cofactor matrix part incorrect: $Co-factor=\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$ I have applied this to D to get: $D_{cofactor}=\begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$ $\endgroup$ – George Jan 18 '18 at 15:49
  • $\begingroup$ As a rule of thumb, for 2x2 matrices switch the diagonal elements and change the sign of the other two. $\endgroup$ – Arnaud Mortier Jan 18 '18 at 15:50
  • $\begingroup$ oh ok, so it means I change the sign to a plus, rather than just applying it in front of it? $\endgroup$ – George Jan 18 '18 at 15:52
  • $\begingroup$ I have again re-attempted to calculate the inverse matrix of D, by your method: I have flipped the signs of the diagonal elements to get: $D_{cofactor}=\begin{bmatrix} - \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}$. This leads to the correct proof. However I don't understand why I can just flip the signs. I was under the impression that to get the cofactor matrix, I simply 'overlay' the following matrix $Co-factor=\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$ with D $\endgroup$ – George Jan 18 '18 at 15:56
  • 1
    $\begingroup$ The Adj operation is the composition of three things: replace every entry by the opposite minor (which here is the opposite entry), multiply by the alternating signs with the pattern that you indicated, and then transpose. $\endgroup$ – Arnaud Mortier Jan 18 '18 at 16:23

Problem #1 and #2, your $AdjD$ step has a mistake, all signs should be negative except for the bottom right


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