İs it true if $0$ is an eigenvalue of a matrix $A$ then $\det{(A)}= 0$ ?
Is there any explanation?
İs it true if $0$ is an eigenvalue of a matrix $A$ then $\det{A}=0$?
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linear-algebra
matrices
eigenvalues-eigenvectors
determinant
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0Just go through the definition of eigenvalue, it's obvious. – 2017-01-18
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0For any matrix $A$ with eigenvalues $\lambda_1,\ldots,\lambda_n$, $\det A = \prod_{i=1}^n \lambda_i$ – 2017-01-18
3 Answers
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Because $|A-\lambda E|=0$ for $\lambda=0$.
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0Isn't equal to detA – 2017-01-18
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$0$ is an eigenvalue iff $\exists x\neq 0\, \, Ax=0\cdot x=0$. This means that $\ker{A}\neq\{0\}$. The linear application represented by $A$ is not injective and $\det{A}=0$
0
When $A$ has eigenvalue $0$, that means there exists a vector $X$ for which $AX=0$.
This is a homogenous linear system, which can only have a non-trivial solution when $\det(A)=0$, because otherwise there would be a solution $A^{-1}AX=A^{-1}\cdot0$ and that implies that $X=0$
But because $0$ is an eigenvalue, we know that there exists a non-zero vector $X$ for which $AX=0$ and thus $\det(A)=0$