A is $4\times 4$ singular matrix. $\operatorname{rank}(A+2I) = 2$ and $\det(A-2I) = 0$

what is the Matrix Characteristic Polynomial of A? is A diagonalizable?

how to solve it or can I get a clue? thanks


closed as off-topic by Theo Bendit, Leucippus, Shailesh, Cesareo, José Carlos Santos Jan 13 at 13:29

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  • $\begingroup$ Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments. $\endgroup$ – saulspatz Jan 12 at 23:28
  • $\begingroup$ @amd, rank. sorry, fixed $\endgroup$ – user3523226 Jan 12 at 23:32
  • $\begingroup$ @amd, soory my mistake. I edited my answer $\endgroup$ – user3523226 Jan 12 at 23:35
  • $\begingroup$ You know three eigenvalues of $A$ and also the geometric multiplicity of one of them. Use that $1+1+2=4$. $\endgroup$ – egreg Jan 12 at 23:42
  • $\begingroup$ I know it and still can't get it.. $\endgroup$ – user3523226 Jan 12 at 23:45

By definition, a number $\lambda$ is an eigenvalue of the matrix $A$ if and only if $A-\lambda I$ is singular (also known as noninvertible).

Therefore, a number $\lambda$ is an eigenvalue of $A$ if and only if $\det(A-\lambda I)=0$.

The dimension of the null space of $A-\lambda I$ is called the geometric multiplicity of $\lambda$. Note that, by definition, the geometric multiplicity is $\ge1$.

The multiplicity of $\lambda$ as a root of the characteristic polynomial $\det(A-XI)$ of $A$ is called the algebraic multiplicity of $\lambda$. So, by definition, if the algebraic multiplicity is $m$, then $(\lambda-X)^m$ divides the characteristic polynomial, but $(\lambda-X)^{m+1}$ doesn't.

Important property. For each eigenvalue of $A$ the algebraic multiplicity is greater than or equal to the geometric multiplicity.

In your case you know three eigenvalues: $\lambda_1=0$, $\lambda_2=2$ and $\lambda_3=-2$. This might not be the complete list, but we can draw some consequences and see.

Denote by $d_i$ and $m_i$ the geometric and algebraic multiplicity of $\lambda_i$.

By the rank-nullity theorem, the geometric multiplicity $d_3=2$. Thus you have $$ 1\le d_1\le m_1, \qquad 1\le d_2\le m_2, \qquad 2=d_3\le m_3 $$ Therefore $$ 1+1+2\le d_1+d_2+d_3\le m_1+m_2+m_3 $$ You should be able to continue, because you know the degree of the characteristic polynomial, don't you?

  • $\begingroup$ Thank you very much $\endgroup$ – user3523226 Jan 16 at 10:07

Essentially, a characteristic polynomial of a matrix is one whose roots are the eigenvalues of that matrix.

We know that a scalar $\lambda$ is an eigenvalue of $A$ if for some non-zero vector $v$ it satisfies $$Av = \lambda v$$ or in other words $$(A-\lambda I)v = 0$$ Even though $v$ is non-zero, left hand multiplication bv $(A-\lambda I)$ results in $0$, so $(A-\lambda I )$ must be singular. This means to find the eigenvalues of $A$ we must find scalars $\lambda$ such that $(A-\lambda I)$ is singular.

For instance, $A$ is singular so $(A-0I)$ is singular.

$det(A-2I)=0$ so $(A-2I)$ is singular by definition.

A square $n\times n$ matrix can only be non-singular if its rank is equal to $n$. This means that $(A+2I)$ is also singular. Furthermore, this matrix has rank $2$, in other words, geometric multiplicity of $2$. This means there are at least two linearly independent eigenvectors associated with this matrix.

In conclusion, we found three different eigenvalues. Each one is guaranteed to produce at least one eigenvector, but we know for a fact that one eigenvalue produces two linearly independent eigenvectors. This makes for a total of four linearly independent eigenvectors. It turns out that since $A$ is a $4\times 4$ matrix it can only have at most four linearly independent eigenvectors. This means we found all of the eigenvalues.

  • $\begingroup$ Thank you for pointing this out @amd . I have edited my answer to make this fact more clear. $\endgroup$ – Ryan Greyling Jan 13 at 6:11
  • $\begingroup$ Thank you very much $\endgroup$ – user3523226 Jan 16 at 10:07

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