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Please forgive my ignorance.

I am busy with a first year course in elementary linear algebra and there are some concepts I do not grasp. Particularly, questions regarding matrix invertibility.

For example, given that

$A^n = 0$ where $n \geq 1$

show that $A$ is not invertible, if $A$ was a matrix.

Now, from what I have learned from web searches is that a when a matrix is equal to $0$, it is call a nilpotent matrix. The problem is that this (including eigenvalues) are not covered in this course, so I am unsure of how to prove the above. Note that the question also doesn't mention $A$ being a square matrix, so I am unsure if even is a nilpotent matrix.

Any small hint to point me in the right direction will be appreciated.

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  • $\begingroup$ $A$ has to be a square matrix so that it is possible to multiply it by itself. $\endgroup$ – Null Sep 9 '14 at 17:02
  • $\begingroup$ DrDeanification: Please see my comment under Arkamis' answer. $\endgroup$ – Michael Hardy Sep 9 '14 at 17:47
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$$A^n = A \cdot A \cdots A\\ \det A^n = \det A \cdot \det A \cdots \det A = \left(\det A\right)^n \\ \det 0 = 0.$$

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  • $\begingroup$ This is an obvious point if one knows about determinants, and certainly one should know about it, but it is an instance of the fact that sometimes a powerful tool that demolishes a problem in an instant doesn't yield as much insight as a more straightforward way to do it. Accordingly, I've added an answer that takes a different approach. $\endgroup$ – Michael Hardy Sep 9 '14 at 17:46
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Suppose $AB=BA = I$. Suppose $n$ is the smallest exponent for which $A^n=0$. $$ 0=B0=BA^n = (BA)A^{n-1} = A^{n-1}. $$ But this contradicts the assumption that $n$ is the smallest.

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Hint: Supposing $A$ is a square $n\times n$ matrix (which we note so the product $A\cdot A=A^2$ makes sense), $A^n$ is invertible exactly when $A$ is.

But if $A^n=0$ it's clear we cannot multiply any matrix $B$ by $A^n$ in order to get the $n\times n$ identity matrix $I$. So $A$ must not be either.

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  • $\begingroup$ I changed $nxn$ to $n\times n$ and $A*A$ to $A\cdot A$. MathJax is based on the way mathematical notation is rendered in $\TeX$, which was invented by Donald Knuth and not by a primitive caveman. $\endgroup$ – Michael Hardy Sep 9 '14 at 17:49
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Suppose $A$ has the inverse $B$, so that $AB=I$, $A^2B^2=AABB=AIB=AB=I$ ... generalise and derive a contradiction.

Note: if you are told that the inverse exists first give it a name, and second, do something with it. What should you do? Use the property of being an inverse in some way. i.e. read the question, and use the key words as clues.

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Hint: Suppose by contradiction that $A$ is invertible. That means that for any $y$, there is an $x$ such that $Ax=y$ and $x=A^{-1}y$. Apply $A^n$ to both sides. You get $0=A^{n-1}y$. Notice that $y$ is arbitrary, then $A^{n-1}=0$. Can you see what this leads to?

Another way of thinking about is that if for a fixed $y$, $x_1$ solves $Ax_1=y$, you can then find an $x_2$ such that $Ax_2=x_1$, so $A^2x_2=y$. Then there is an $x_3$ such that $A^3x_3=y$, etc. This will also lead to a contradiction.

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