Does my proof hold up to prove that \$0\$ is the only eigenvalue of \$A\$ if \$A^2 = 0\$?

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.

You are watching: For a to have 0 as an eigenvalue, k must be

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\$.

See more: The Primary Objective Of Financial Accounting Information Is To Provide Useful Information To: 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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