Let $A$ be an $n imes n$ matrix.$A^2 = A imes A$ because of matrix multiplication.If $A = k$, where $k
eq 0$, then $A^2
eq 0$. Thus if $A^2 = 0$, $A$ is the zero matrix.
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If $A$ is the zero matrix, det($A^2) = 0$.$lambda eq 0$ if $A$ is non-singular. $A$ is a non-singular matrix IFF $det(A)$ $ eq 0$. The charcteristic equation is false when $lambda = 0$ because that a non-singular matrix so it can only it is in true when the matrix is singular, make $det(A) = 0$.$lambda = 0$,$ ightarrow$$eginvmatrix0I - 0endvmatrix= 0$
It is wrong. Friend seem to think that if $A eq0$, climate $A^2 eq0$, i beg your pardon is false. Take it $A=left(eginsmallmatrix0&1\0&0endsmallmatrix ight)$, for instance.
If $A$ had an eigenvalue $lambda eq0$, climate there would be a vector $v eq0$ such that $A.v=lambda v$. So,eginalignA^2.v&=A.(A.v)\&=A(lambda v)\&=lambda(A.v)\&=lambda^2v\& eq0endalignand therefore $A^2 eq0$.
What you wrote is a mess.
If $A = k$, wherein $k eq 0$, then $A^2 eq 0$. Therefore if $A^2 = 0$, $A$ is the zero matrix.
This is false: think about $A=eginpmatrix0&1\0&0endpmatrix$.
The rest I cannot also comment on.
What you desire to perform is the following. Expect there is one eigenvalue $lambda eq 0$. Then there is an eigenvector $vin hedesigningfairy.combb C^nsetminus $ such that $Av=lambda v$. Yet then $A^2v=lambda^2v eq 0$, therefore the matrix $A^2$ cannot be $0$.
There are non-zero matrices $A$ whereby $A^2 = 0$ so your proof go not hold true. If $A^2 = 0$, we currently know that $A$ is non-invertible since if $A$ to be invertible, we might multiply both sides through $A^-1$ and also get $A = 0$, yet $0$ is not invertible. Us do know $0$ is an eigenvalue the $A$ though due to the fact that we deserve to multiply $A$ with any column that $A$ to attain $0$. We now have actually to present $0$ is the just eigenvalue. Mean there is an eigenvalue $lambda eq 0$. Then, $Ax = lambda x$ and also $0 = A^2x = AAx = Alambda x = lambda Ax$. Due to the fact that $lambda eq 0$, we acquire $Ax = 0 = 0x$ for part eigenvector $x$, yet this method $0$ is one eigenvalue for $x$, i beg your pardon contradicts $lambda eq 0$. Thus, $lambda = 0$ is the just eigenvalue because that $A$.
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Here is an alternative proof: $x^2$ is an annihilating polynomial for the matrix $A$, for this reason the minimal polynomial because that $A, m_A(x)$ divides $x^2implies m_A(x)=x, x^2$. $0$ is the just root that the minimal polynomial, so $A$ just has actually one distinct eigenvalue: $0$.
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Queries in the proof of a square procession $A$ is invertible if and also only if $lambda = 0$ is not an eigenvalue of $A$
carry out a proof or counterexample. If $M$ is an $n imes n$ procession that is diagonal, then every non-zero worth in $M$ is one eigenvalue of $M$.
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