Question

Okay, let's try a different question. Still linear operators.

(a) Let A=\(\left[\begin{matrix}a & b\\c&d\end{matrix}\right]\) be a real matrix, with c not zero. Show that using conjugation by elementary matrices, one can eliminate the "a" entry.

(b) Which matrices with c=0 are similar to a matrix in which the "a" entry is zero?

Answer 1

I already solved (a), with the elementary matrix in question being \[
\left[\begin{matrix}1&ac^{-1}\\0&1\end{matrix}\right]
\]But, I'm not sure about part (b).

Answer 2

Clarifying points: Conjugating means doing \(E^{-1}AE\). The matrix above in my solution is actually the \(E^{-1}\). Similar means that there is a matrix you can conjugate by to reach the other matrix.

Answer 3

I'm thinking that the answer is all singular matrices, but I'm not sure if that's right and if it is I don't know how to prove it

Answer 4

It might be so simple as just to be all matrices where a is also zero. Still not sure how to prove that though.

Answer 5

wouldn't the answer to the first one be:\[\left[\begin{matrix}1&-ac^{-1}\\0&1\end{matrix}\right]\]

Answer 6

Yes asna, that would be E, sorry I was unclear there, the one I put was E inverse.

Answer 7

and for the second one, if you left-multiply the matrix with a transformation that just swaps the two rows, then they would be similar if all the non-zero elements are in the same ratio - I /think/

Answer 8

If you conjugate by a permutation (row-swap) matrix it just sends it back to where it started. We have to conjugate, we can't just left-multiply.

Answer 9

i.e.:\[\left[\begin{matrix}0 & b_1\\c_1&d_1\end{matrix}\right]\]is similar to:\[\left[\begin{matrix}a_2 & b_2\\0&d_2\end{matrix}\right]\]if:\[\frac{b_1}{d_2}=\frac{d_1}{b_2}=\frac{c_1}{a_2}\]

Answer 10

I lost you there.

Answer 11

which I believe is the same as saying they are similar if:\[\left[\begin{matrix}0 & b_1\\c_1&d_1\end{matrix}\right]=k\times \left[\begin{matrix}0 & 1\\1&0\end{matrix}\right]\left[\begin{matrix}a_2 & b_2\\0&d_2\end{matrix}\right]\]

Answer 12

where k is some constant

Answer 13

Similar here means that they are conjugate to each other. Meaning, A is similar to B if \(A=P^{-1}BP\) for some invertible matrix P.

Answer 14

ok - I was looking at from the point of view of solving simultaneous equations, so the two would be similar if one was just a multiple of the other. I guess that is not what is being asked here.

Answer 15

Yeah, not quite the same. We're looking here at similar with respect to conjugation because that is the relevant equivalence when looking at changes of basis.

Answer 16

Basically, we're looking at it as a linear transformation, and keeping the transformation the same, but changing the basis of the vector space that the transformation is applied to.

Answer 17

so we are saying that:\[\left[\begin{matrix}0 & b_1\\c_1&d_1\end{matrix}\right]=\left[\begin{matrix}1 & t\\0&1\end{matrix}\right]\left[\begin{matrix}a_2 & b_2\\0&d_2\end{matrix}\right]\left[\begin{matrix}1 & -t\\0&1\end{matrix}\right]\]correct?

Answer 18

Yes, unless there is some other form of matrix that you could conjugate by that would remove the a (I don't think there is, but I don't know).

Answer 19

So you see why I'm thinking that a has to be zero as well?

Answer 20

this works out to be:\[\left[\begin{matrix}0 & b_1\\c_1&d_1\end{matrix}\right]=\left[\begin{matrix}a_2 & b_2+(d_2-a_2)t\\0&d_2\end{matrix}\right]\]

Answer 21

which would indeed imply that \(a_2=0\), but also that \(c_1=0\) and \(d_1=d_2\) and:\[b_1=b_2+(d_2-a_2)t\]

Answer 22

And similarity means you have to conjugate by elementary matrices, and no other type of elementary matrix would be able to do this... so a does have to equal zero. That's a satisfying proof for me. Thanks.

Answer 23

is there any online material on this specific topic?
I would be interested in learning more about it.

Answer 24

Hrm, I am not sure. This is from Artin's Algebra. I know that is the book used in the Harvard online course in Algebra taught by Benedict Gross, but I haven't watched enough lectures to know if he covers this specific stuff much.

Answer 25

ok - thanks for info in any case.

Answer 26

"[. . .]similarity means you have to conjugate by elementary matrices."

I don't agree with this. What is your reasoning?

Answer 27

That's by definition.

Answer 28

Not from where I'm looking it up. http://en.wikipedia.org/wiki/Similar_matrix

"for some invertible n-by-n matrix P"
"with P being the change of basis matrix" <-- Is this perhaps the basis for your statement, pun unintended?

Answer 29

An invertible nxn matrix is by necessity a product of elementary matrices.

Answer 30

Why? I'm not exactly seeing this. Could you show me this, for example, with \(
\begin{bmatrix}
1 & 2\\
3 & 4
\end{bmatrix}
?\)

Answer 31

hmmm... yes indeed, according to that link posted by @Limitless P does not have to an elementary conjugate matrix

Answer 32

there are some special cases that it mentions:

"In the definition of similarity, if the matrix P can be chosen to be a permutation matrix then A and B are permutation-similar; if P can be chosen to be a unitary matrix then A and B are unitarily equivalent."

Answer 33

I have no idea why decomposing that matrix into elementary matrices is taking me so long. This is weird. Regardless, it is true that every invertible matrix is a product of elementary matrices.

Answer 34

\[
\begin{bmatrix}
1 & 2\\
3 & 4
\end{bmatrix}=\begin{bmatrix}
1 & 0\\
3 & 1
\end{bmatrix}\begin{bmatrix}
1 & -1\\
0 & 1
\end{bmatrix}\begin{bmatrix}
1 & 0\\
0 & -2
\end{bmatrix}
\]There we go. Took me long enough.

Answer 35

The intuition is quite easy on that. If it's invertible then you can reduce it to the identity by a series of elementary operations. So, you must also be able to build it from the identity by a series of elementary operations.

Answer 36

but:\[\begin{bmatrix}
1 & 0\\
3 & 1
\end{bmatrix}\ne\begin{bmatrix}
1 & 0\\
0 & -2
\end{bmatrix}^{-1}\]

Answer 37

I'm not conjugating there, I'm showing Limit that the matrix he provided was a product of elementary matrices.

Answer 38

so its not of the form:\[B=P^{-1}AP\]to show similarity

Answer 39

Again, I'm not showing similarity there. I'm showing Limitless why every invertible matrix is a product of elementary matrices.

Answer 40

oh - sorry - I should learn to read! :D

Answer 41

No worries :) So, since every invertible matrix is a product of elementary matrices, that means that conjugating must be by elementary matrices. If you conjugate multiple times you remain similar, so you can use any invertible matrix, but you can always break it down into elementary matrices.

Answer 42

Rather interesting question we have here, guys. Thanks.

Answer 43

I'm still very, very iffy on your reasoning, @nbouscal. But, I think it is correct as I can't pose any objections at the moment.

Answer 44

Look at it this way. If A is invertible, so is the end result A' of its row reduction. Since an invertible matrix cannot have a row of zeros, and since a square row echelon matrix is either the identity or has a row of zeros, A' must be the identity. If A can be reduced to the identity by a sequence of elementary row operations, then A must be a product of elementary matrices, because elementary matrices have elementary inverses.

Answer 45

Does that help? I'm not sure how far to break this down.

Answer 46

I'm referring to your proof as a whole with respect to part (b).

It's just not setting well with me for some reason.