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Suppose that $\det\left(A-(1+\beta)I\right) = 0$ where $A$ is invertible matrix and $\beta$ is some positive real number. Can the biggest eigenvalue of $A$ smaller than $1+\beta$?

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    $1+\beta$ is an eigenvalue, so the biggest eigenvalue is at least $1+\beta$.2012-12-16
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    @MichaelAlbanese Why does $\det(A - \lambda I)=0$ imply $Ax = \lambda x$ for all suitable column vectors $x$ ?2012-12-16
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    $\det B=0$ means $B$ has non-trivial kernel, so $Bx=0$ for some non-zero column vector $x$. Now apply this to $B=A-\lambda I$, multiply out the brackets and rearrange.2012-12-16
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    @user108903 Thanks for that! I feel a bit silly now.2012-12-16

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Well you're assuming that $1+ \beta$ is an eigenvalue, so (assuming all the eigenvalues are real), certainly the largest one is larger than this.