Similar matrices

  • Thread starter ehrenfest
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  • #1
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Homework Statement


1) Show that for any real numbers a and b, the matrices

[tex] \left( \begin{array}{cc}
1 & a \\
0 & 2 \\
\end{array}\right)[/tex]

and

[tex] \left(\begin{array}{cc}
1 & b \\
0 & 2 \\
\end{array}\right)[/tex]

are similar.

2) Show that

[tex] \left( \begin{array}{cc}
2 & 1 \\
0 & 2 \\
\end{array}\right)[/tex]

and

[tex] \left(\begin{array}{cc}
2 & 0 \\
0 & 2 \\
\end{array}\right)[/tex]

are not similar.

Homework Equations





The Attempt at a Solution



We want to show that B=P^{-1} A P holds for some P in the first case and holds for no P in the second case. So I let P be an arbitrary 2 by 2 matrix and just wrote out the four equations that you get using the explicit formula for the inverse but that failed . So what is the trick...
 

Answers and Replies

  • #2
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anyone?
 
  • #3
HallsofIvy
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There is no "trick", there is only algebra! You say you wrote out the four equations, using an arbitrary (invertible?) 2 by 2 matrix. That should work. Exactly what did you get?

By the way, if B= P^{-1}AP, then, multiplying both sides, on the left, by P gives PB= AP. That might be simpler to use for an "arbitrary 2 by 2 matrix".
 
  • #4
Dick
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Might be a good idea to look at the eigenvectors of each matrix.
 
  • #5
HallsofIvy
Science Advisor
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True, but:
1. Since Ehrenfest said he had already tried an elementary way that should work, I thought it best to encourage him in that. He may not know about eigenvalues yet.

2. While two matrices having different eigenvalues cannot be similar, having the same eigenvalues does not necessarily mean they are similar.
 
  • #6
2,012
1
There is no "trick", there is only algebra! You say you wrote out the four equations, using an arbitrary (invertible?) 2 by 2 matrix. That should work. Exactly what did you get?

By the way, if B= P^{-1}AP, then, multiplying both sides, on the left, by P gives PB= AP. That might be simpler to use for an "arbitrary 2 by 2 matrix".
OK. Here is the solution to part 2: Let
P = [tex]
\left( \begin{array}{cc}
w & x \\
y & z \\
\end{array}\right)
[/tex]

Then, if we write equations for each element of AP=PB, where A is the first matrix and B is the second matrix, we see that w=y=0 and a matrix with a column of zeros is clearly not invertible.

We can do the same thing for part 1) and obtain PA=BP (where A is the first matrix and B is the second one) if x=y=0 and w=b and z=a. Then P is invertible whenever a or b are not zero. If they are both zero, this is trivial. If a is zero and b is nonzero we let x=b,z=1, and w= anything nonzero, y = 0. And then when a is nonzero and b is zero do the "same" thing.
 
Last edited:

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