Ask your own question, for FREE!
Mathematics 61 Online
OpenStudy (anonymous):

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?

OpenStudy (anonymous):

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).

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

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

OpenStudy (asnaseer):

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/

OpenStudy (anonymous):

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.

OpenStudy (asnaseer):

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:\[\frac{b_1}{d_2}=\frac{d_1}{b_2}=\frac{c_1}{a_2}\]

OpenStudy (anonymous):

I lost you there.

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

where k is some constant

OpenStudy (anonymous):

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.

OpenStudy (asnaseer):

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.

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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.

OpenStudy (asnaseer):

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?

OpenStudy (anonymous):

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).

OpenStudy (anonymous):

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

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

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.

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

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.

OpenStudy (asnaseer):

ok - thanks for info in any case.

OpenStudy (anonymous):

"[. . .]similarity means you have to conjugate by elementary matrices." I don't agree with this. What is your reasoning?

OpenStudy (anonymous):

That's by definition.

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (asnaseer):

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

OpenStudy (asnaseer):

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."

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

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

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

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

OpenStudy (asnaseer):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

Rather interesting question we have here, guys. Thanks.

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Latest Questions
danielfootball123: What is the name of the president who got assassinated?
13 minutes ago 2 Replies 0 Medals
MAGABACK: ART!
3 hours ago 5 Replies 0 Medals
danielfootball123: Is Donald trump a good president?
5 hours ago 92 Replies 6 Medals
Gucchi: chem help
17 hours ago 9 Replies 0 Medals
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!