Question

Square matrix

Answer 1

If \[\left[ B \right]^2 = \left[\begin{matrix}2 & 3 \\ 0 & 2\end{matrix}\right] \]
find [B] !

Answer 2

My approach is just construct the matrix B with arbitrary numbers, then square it, setting it equal to this matrix. Now you can solve for those arbitrary numbers.

Answer 3

\[\left[ B \right] \left[ B \right] = \left[\begin{matrix}2 & 3 \\ 0 & 2\end{matrix}\right]\]

Answer 4

\[\Large [B]=\left[\begin{matrix}a& b\\ c & d\end{matrix}\right]\] \[\Large [B]^2=\left[\begin{matrix}a^2+bc& ab+db\\ ac+cd & bc+d^2\end{matrix}\right]= \left[\begin{matrix}2 &3\\ 0 & 2\end{matrix}\right]\]

Answer 5

Look at the bottom left equation, we get:

c(a+d)=0

that means either c=0 or a=-d.

However if we look at the top right equation

d(a+b)=3

we know that a can't be -d, otherwise it would say 0=3. So we know c=0.

This simplifies the top left and bottom right equations down to a^2=2 and d^2=2 and from earlier we know that since a isn't -d that a=d and they are either both positive or negative square root of 2. Now we can plug this into the top right equation again to get that and there you go.

It looks like there are two possible answers for what B can be. Not weird though, we're used to having multiple answers to quadratics, it also works for matrices. heh

Answer 6

nice :) power of matrix wont work with repeated eigenvalues is it ?

Answer 7

Woah I didn't even think of that.

Answer 8

im going thru this solution which has distinct eigen values
http://math.stackexchange.com/questions/732511/fractional-power-of-matrix

but our matrix has repeated eighenvalues so i was not able to get an inverse for eigenvector matrix

Answer 9

Are you saying we could diagonalize this matrix and then apply Newton's formula for finding square roots?

Answer 10

yeah \[\large A^n = S\Lambda^n S^{-1}\]

Answer 11

it seems we cannot diagonalize this matrix

Answer 12

atleast its not possible using eigen vectors

Answer 13

I have an idea though

Answer 14

What if we square the matrix? Then we can diagonalize it. Then we just take the square root twice.

Answer 15

OGM Kai !!! lets see if it works xD

Answer 16

squaring gives the same bad matrix :/

Answer 17

yeah eigen vectors don't change when we take powers, so once a matrix is bad, it is bad forever :o

Answer 18

Awww I messed up when I squared it. :,(

Answer 19

But I think we learned something. The only number that changed is the top right number went from a 3 to a 12. So maybe squaring the matrix just multiplies the top right number by 4. So that means the square root should just have a 3/4 there.

Sure enough, it works!

Answer 20

In order for it to be a 2x2 defective matrix, we have to have \[\LARGE \left[\begin{matrix}a & x \\ 0 & a\end{matrix}\right]\] because repeated eigenvalues come from \[\LARGE (a- \lambda )^2=0\]so that means squaring brings us to \[\LARGE \left[\begin{matrix}a & x \\ 0 & a\end{matrix}\right]^2=\left[\begin{matrix}a & 2ax \\ 0 & a\end{matrix}\right]\] in general.

Answer 21

I see..

a_21 stays fixed at 0
the diagonal values give us eigenvalies, so we just take sqrt of them ?
and a_12 = x/2a

sweet <3

Answer 22

No it's not square rooting, we just plug in. Here we have 2ax=3 and a=2. That means when we solve for x, we have x=3/4.

Then there's one added extra. We could multiply the matrix by a scalar, -1. That will go away when we square it too. =P

Answer 23

or maybe I read your response wrong idk

Answer 24

\[\LARGE \left[\begin{matrix}a & x \\ 0 & a\end{matrix}\right]^2=\left[\begin{matrix}a^2 & 2ax \\ 0 & a^2\end{matrix}\right] \]

Answer 25

Nope, that's wrong.

Answer 26

we need to take sqrt along diagonal to construct A back from A^2 right ?

Answer 27

No we can't diagonalize this matrix, this is just a general form of the matrix we started out with.