Let $K=(A+D)^{-1}A$ where $A$ is symmetric positive definite and $D$ is a diagonal matrix with positive elements. Is it true that $\|K\|\leq 1$ where $\|\cdot\|$ is the induced $2$-norm?
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What do you mean by the $2$-induced norm? The square root of the sum of squares of the entries of the matrix is one possible interpretation; the norm of the matrix as an operator between $l_2^n$ spaces is another; which one is it? – uniquesolution Dec 06 '17 at 13:36
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1@uniquesolution it's a standard term for the second one. That is, he means the "spectral norm". – Ben Grossmann Dec 06 '17 at 13:38
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$|\cdot|_2$ denotes the maximum singular value. – user293017 Dec 06 '17 at 13:40
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@uniquesolution "induced" here refers to the fact that the matrix norm is derived from the vector norm, i.e. we compute the norm of the operator between vector spaces. – Ben Grossmann Dec 06 '17 at 13:41
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@user293017 you were right about that; good catch – Ben Grossmann Dec 06 '17 at 14:03
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If $AD=DA$, the inequality is true: you have, if $d_n$ is the least entry in the diagonal of $D$, $$ A+D\geq A+d_nI. $$ So $(A+D)^{-1}\leq(A+d_nI)^{-1}$, and so $$ (A+D)^{-1}A=A^{1/2}(A+D)^{-1}A^{1/2}\leq A^{1/2}(A+d_nI)^{-1}A^{1/2}=(A+d_nI)^{-1}A $$ Now the inequality follows from the fact that inequalities between positive elements preserve norm, and $(A+d_nI)^{-1}A$ is positive and has eigenvalues $\lambda/(\lambda+d_n)$.
In general, the answer is no. For instance with $$ A=\begin{bmatrix} 2&1\\1&1\end{bmatrix},\ \ D=\begin{bmatrix} 1&0\\0&2\end{bmatrix}, $$ we have $$ K=(A+D)^{-1}A=\frac13\,\begin{bmatrix}5&4\\7&4\end{bmatrix}. $$ Then $\|K\|\geq 7/3>1$.
Martin Argerami
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Thank you for your response! Why $(A+D)^{-1}A=A^{1/2}(A+D)^{-1}A^{1/2}$? – user293017 Dec 06 '17 at 15:47
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