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I have the matrix

$$A =\begin{pmatrix}0&1\\1&0\end{pmatrix}$$

How would I diagonalize it using elementary row operations? It's been a while since I've worked with them so I'm doubting myself when doing the same operations to both columns and rows. Steps would be much appreciated.

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You diagonalize a matrix by finding eigenvalues and eigenvectors, not by using elementary row operations. –  Gerry Myerson Feb 14 '13 at 1:09
    
Yes but my professor told us explicitly he wants us to do it that way. If this helps, we are trying to find a matrix B that is congruent to A –  tamefoxes Feb 14 '13 at 1:15
    
Also, the field is not specified so we can't assume complex eigenvalues exist –  tamefoxes Feb 14 '13 at 1:15
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I know what it means to find a matrix that is similar to another; I don't know what it means to find a matrix that is congruent to another. Also, the eigenvalues are $\pm1$, which are there no matter what field you are working in (although things may be a little tricky in fields of characteristic $2$). –  Gerry Myerson Feb 14 '13 at 2:20
    
@GerryMyerson Here is a definition mathworld.wolfram.com/CongruentMatrices.html But like you, I don't see what this has to do with diagonalization. –  1015 Feb 14 '13 at 3:16

2 Answers 2

up vote 1 down vote accepted

I note that the question refers to "doing the same operations to both columns and rows," so maybe this is what's intended.

Starting with $$\pmatrix{0&1\cr1&0\cr}$$ add the 2nd row to the 1st, then add the 2nd column to the 1st; you get $$\pmatrix{1&1\cr1&0\cr}{\rm\ then\ }\pmatrix{2&1\cr1&0\cr}$$ Now subtract half the 1st row from the 2nd, followed by subtracting half the 1st column from the second; $$\pmatrix{2&1\cr0&-1/2\cr}{\rm\ then\ }\pmatrix{2&0\cr0&-1/2\cr}$$ and there's your diagonal matrix.

Now, adding $a$ times the 2nd row to the 1st is the same as multiplying on the left by $$\pmatrix{1&a\cr0&1\cr}$$ and adding $a$ times the 2nd column to the 1st is the same as multiplying on the right by the transpose, $$\pmatrix{1&0\cr a&1\cr}$$ So we really are getting $A=P^tBP$ with $B$ diagonal, and you can walk through the steps to see what $P$ is.

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Yes this is exactly what I was looking for, I didn't know how else to put it. Sorry for the lack of proper terminology Thanks a lot! –  tamefoxes Feb 14 '13 at 6:25

Note that your matrix is symmetric, so you know from the start that there exists an orthogonal matrix $P$ such that $P^tAP=P^{-1}AP$ is diagonal.

Now compute the determinant of $XI-A$ to find the characteristic polynomial of $A$. You'll find $$ p_A(X)=X^2-1=(X-1)(X+1). $$

So the eigenvalues are $-1$ and $+1$.

To look for an eigenvector associated with $1$, you must solve the system $$ A\left(\matrix{x\\y}\right)=\left(\matrix{x\\y}\right). $$ There are not so many elementary row operations to perform to find that $x=y$. To get a unit vector such that $x=y$, we take $(1/\sqrt{2},1\sqrt{2})$.

Doing the same with the eigenvalue $-1$, we find $(1/\sqrt{2},-1\sqrt{2})$.

Now set $$ P:=\left( \matrix{\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}}}\right). $$

You can check, you have $$ P^tAP=P^{-1}AP=\left(\matrix{1&0\\0&-1}\right). $$

So $A$ and this diagonal matrix are similar, and congruent.

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Thanks for the help, if anyone can find a way to "diagonalize" the matrix A by elementary row operations, feel free to post! –  tamefoxes Feb 14 '13 at 4:11
    
@user43956 Your question does not really make sense. You perform elementary row operations to solve linear systems. Here, when you solve for eigenvectors. Have you written down these systems? THey're trivial. THere is nothing to do to solve them. –  1015 Feb 14 '13 at 5:06
    
I am not trying to solve the system, I am trying to diagonalize A into a new matrix B such that A can be expressed as P^TBP where P is an invertible matrix and the product of elementary matrices that diagonalize A by performing elementary row and column operations –  tamefoxes Feb 14 '13 at 20:31

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