Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 16

MATH152 April 2013

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[hide] Question A 16
Let A and B be two $n\times n$ matrices and let v be a non-zero vector in $\mathbb {R} ^{n}$ . Given that $Av=Bv$ , can we conclude that $A=B$ ? Justify your answer.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
If we isolate for 0, we get $\displaystyle 0=Av-Bv=(A-B)v$ so $A=B$ could be true but it seems like there are other cases to consider. See if you can find a counter example.

[show] Hint 2
Watch out for the $n=1$ case as it is an exception. This is the case where you deal with real numbers instead of matrices (real numbers are just $1\times 1$ matrices)

[show] Hint 3
Lastly, to help with the counter example, find matrices A and B so that $A-B$ has nontrivial nullspace.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
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[show] Solution
There is a small subtlety in this problem. Suppose first that $n=1$ . Then $Av=Bv$ becomes an equation of real numbers, as 1x1 matrices are just one number, and a v is also just one number. This gives $(A-B)v=0$ and as $v\neq 0$ (the problem stated this was not zero) we have that $A=B$ . Now, for $\displaystyle n=2$ , suppose that $A={\begin{bmatrix}1&0\\0&0\end{bmatrix}}$ and let $\displaystyle B=-A$ , Then $A-B={\begin{bmatrix}2&0\\0&0\end{bmatrix}}$ Let $v={\begin{bmatrix}0\\1\end{bmatrix}}$ Then this vector is in the nullspace of both matrices, Av = 0 = Bv, but $A\neq B$ and hence the claim is not true for $n=2$ . Next, suppose that n is at least 3. We can generalize the above example nicely. Choose A to be the all zeroes matrix except with a 1 in the position $A_{1,1}$ (the top left corner) and let $B=-A$ so that it too has all zeroes except a -1 in the top left corner. Then $A-B$ has only a 2 in the top left corner and zeroes everywhere else. Lastly, let $v={\begin{bmatrix}0\\0\\\vdots \\0\\1\end{bmatrix}}$ Then notice that $Av=0=Bv$ however $A\neq B$ . Thus, the claim is false for all $n\geq 2$ but is true when $n=1$ .