If A^2 = 0, then A is not an invertible matrix

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If \( A^2 = 0 \), then matrix \( A \) is definitively not invertible. The proof by contradiction shows that assuming \( A \) is invertible leads to the conclusion that \( A = 0 \), which contradicts the definition of an invertible matrix. Additionally, examples of nonzero matrices that satisfy \( A^2 = 0 \), such as the matrix \(\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\), further illustrate that these matrices cannot be invertible.

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Mr Davis 97
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Homework Statement


Suppose that ##A^2 = 0##. Show that ##A## is not an invertible matrix

Homework Equations

The Attempt at a Solution


We can do a proof by contradiction. Assume that ##A^2 = 0## and that ##A## is invertible. This would imply that ##A=0##, which is to say that A is not invertible, since ##0## has no inverse. This is a contraction, so it must be the case that if ##A^2 = 0##, then ##A## is not invertible.

Is this the way I should be doing this problem?
 
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Mr Davis 97 said:

Homework Statement


Suppose that ##A^2 = 0##. Show that ##A## is not an invertible matrix

Homework Equations

The Attempt at a Solution


We can do a proof by contradiction. Assume that ##A^2 = 0## and that ##A## is invertible. This would imply that ##A=0##, which is to say that A is not invertible, since ##0## has no inverse. This is a contraction, so it must be the case that if ##A^2 = 0##, then ##A## is not invertible.

Is this the way I should be doing this problem?
I first thought - and this might well have been intended - that you should show, that there is a non-trivial element in the kernel of ##A##, namely the entire image of ##A##, but I like your solution better.
 
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Mr Davis 97 said:

Homework Statement


Suppose that ##A^2 = 0##. Show that ##A## is not an invertible matrix

Homework Equations

The Attempt at a Solution


We can do a proof by contradiction. Assume that ##A^2 = 0## and that ##A## is invertible. This would imply that ##A=0##, which is to say that A is not invertible, since ##0## has no inverse. This is a contraction, so it must be the case that if ##A^2 = 0##, then ##A## is not invertible.

Is this the way I should be doing this problem?

Looks good to me as well. Note that you can prove this directly by using determinants, but I suspect you are not allowed to use determinants at this stage.
 
Mr Davis 97 said:
We can do a proof by contradiction. Assume that ##A^2 = 0## and that ##A## is invertible. This would imply that ##A=0##,
There are nonzero matrices so as A2=0. You should prove that they are not invertible.
For example, the square of the following matrix is zero.
\begin{pmatrix}

0 & 1 \\
0 & 0

\end{pmatrix}
 
Last edited:
ehild said:
There are nonzero matrices so as A2=0. You should prove that they are not invertible.

If A^2 = 0 and A is invertible, this implies A^(-1) A^2 = A^(-1) 0 = 0. No need to bother with non-invertible A's here.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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