For two p.d. matrices $A$ and $B$, prove that $\lambda_1(AB)\leqslant \lambda_1(A) \cdot\lambda_1(B)$

Question

If $A$ and $B$ are two nxn positive definite matrices, then show that

$$\lambda_1(AB) \leqslant \lambda_1(A) \cdot \lambda_1(B),$$

where $\lambda_1(\cdot)$ denotes the largest eigenvalue.

Answer 1

Let the eigen values of $AB$ be $c_1, c_2, \ldots,c_n$ and those of $A$ and $B$ be $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ respectively. Now, we know that, $\det(A)$ is the product of the eigenvalues of $A$, counted with their algebraic multiplicities.

Also, we have, $\det(AB) = \det(A)\det(B)$

$$ \Rightarrow c_1 c_2 \cdots c_n = (a_1 a_2 \cdots a_n)(b_1 b_2 \cdots b_n) $$

=> $c_1$ * $c_2$ * ... * $c_n$ = ($a_1$ * $a_2$ * ... * $a_n$) * ($b_1$ * $b_2$ * ... * $b_n$ )

Using the above equality/identity(since all eigenvalues are positive, $A$ & $B$ being p.d.), we can evidently write:

$$\lambda_1(AB) \le \lambda_1(A) \lambda_1(B).$$

Answer 2

Let $C$ be a positive symmetric $n\times n$ matrix, and let $|\cdot|$ be the standard euclidean norm on $\Bbb R^n$. For all $X\in\Bbb R^n$, $$|CX|\leq\lambda_1(C)|X|$$ with equality iff $X$ is eigenvector of $C$ with eigenvalue $\lambda_1(C)$, so that, again, for $C$ positive symmetric, $$\lambda_1(C)=\max_{|X|\leq 1}|CX|=\|C\|$$ where $\|\cdot\|$ is the matrix norm subordinate to the standard euclidean norm on $\Bbb R^n$. The answer to your question now follows, as for any matrix $D$ and any eigenvalue $\lambda$, $|\lambda|\leq \|D\|$ so that $$|\lambda_1(AB)|\leq\|AB\|\leq\|A\|\cdot\|B\|=\lambda_1(A)\lambda_1(B)\,.$$

Answer 3

$AB$ is similar to an SPD matrix, hence they have same eigenvalues: $$ AB\sim B^{1/2}(AB)B^{-1/2}\quad\Rightarrow\quad\lambda_{1}(AB)=\lambda_1(B^{1/2}AB^{1/2}). $$ Using this and the variational characterisation of $\lambda_1$, we have $$ \begin{split} \lambda_1(AB)&=\max_x\frac{x^TB^{1/2}AB^{1/2}x}{x^Tx} =\max_y\frac{y^TAy}{y^TB^{-1}y} =\max_y\frac{y^TAy}{y^Ty}\frac{y^Ty}{y^TB^{-1}y}\\ &\leq\max_y\left(\frac{y^TAy}{y^Ty}\right)\left(\max_y\frac{y^Ty}{y^TB^{-1}y}\right) =\left(\max_y\frac{y^TAy}{y^Ty}\right)\left(\max_z\frac{z^TBz}{z^Tz}\right) =\lambda_1(A)\lambda_1(B). \end{split} $$