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Looking at the vector space $\Bbb C^n$ with the standart inner product, Let $v_1,...v_n \in \Bbb C^n$ and $A \in M_n(\Bbb C)$ a matrix with columns $v_1,...,v_n$. Prove that:

$$ \lvert detA\rvert \leq \prod_{j=1}^n ||v_j||$$

Furthermore, prove that equality holds if and only if $v_1,...,v_n$ is an orthogonal sequence.

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2 Answers 2

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If $\operatorname{rank} A < n$ then $\det(A) = 0$ and the inequality is clear. In this case, the equality holds iff $v_i = 0$ for some $1 \leq i \leq n$ (note that this doesn't mean that $v_1,\dots,v_n$ is an orthogonal sequence).

On the other hand, if $A$ has full rank, perform a QR decomposition and write $A = QR$ when $Q$ is unitary and $R$ is upper triangular. The columns $e_1,\dots,e_n$ of $Q$ are the result of performing the Gram-Schmidt procedure on $v_1,\dots,v_n$ while the diagonal entries of $R$ are $\left< v_i, e_i \right>$. Hence,

$$ |\det(A)| = |\det(QR)| = |\det(Q)||\det(R)| = |\det(R)| = \prod_{i=1}^n |\left< v_i, e_i \right>| \leq \prod_{i=1}^n \| v_i \| \| e_i \| = \prod_{i=1}^n \| v_i \|. $$

Equality holds if and only if $| \left< v_i, e_i \right> | = \| v_i \|$ for all $1 \leq i \leq n$. Since equality in the Cauchy-Schwartz inequality holds if and only if the vectors are linearly dependent, we can write $v_i = c_i e_i$ for some $c_i \in \mathbb{C}$ with $\| v_i \| = c_i$ which implies that $(v_i)_{i=1}^n$ are orthogonal because $(e_i)_{i=1}^n$ are.

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  • $\begingroup$ Great proof. Is there another way to justify $ \lvert detA\rvert= \lvert detR\rvert$ other than the QR decomposition? (I am suppose to prove it without using it as I havn't learned this theorem). $\endgroup$
    – user401516
    Jun 12, 2017 at 8:53
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    $\begingroup$ I don't think you'll get something which is easier than just proving it. For square matrices of full rank (which is what you are interested in), the proof follows immediately from the Gram-Schmidt procedure. Namely, the equations $v_1 = \left< v_1, e_1 \right> e_1, v_2 = \left< v_2, e_1 \right> e_1 + \left< v_2, e_2 \right> e_2, \dots, v_n = \left< v_n, e_1 \right> e_1 + \dots + \left< v_n, e_n \right> e_n$ written in matrix form give you the $QR$ decomposition. $\endgroup$
    – levap
    Jun 12, 2017 at 9:05
  • $\begingroup$ Alright, I will try to prove it. But why is it that $ \lvert detQ\rvert= 1$ ? $\endgroup$
    – user401516
    Jun 12, 2017 at 9:16
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    $\begingroup$ A unitary matrix ($QQ^{*} = I$) has determinant of absolute value one because $1 = \det(I) = \det(Q)\det(Q^{*}) = \det(Q) \overline{\det{Q}} = |\det(Q)|^2$ implies $|\det(Q)| = 1$. $\endgroup$
    – levap
    Jun 12, 2017 at 9:20
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Here is more of a comment than a proof, actually a partial proof limited to 3d, where $\vec v_1, \vec v_2,$ and $\vec v_3\in \Bbb R^3$. The determinant of $A$ is the triple scalar product $$\det A=\vec v_1\cdot(\vec v_2\times \vec v_3)$$ Using the Cauchy-Schwartz inequality $$ |\det A|\leq|\vec v_1||\vec v_2\times \vec v_3| $$

Using the Levi-Civita symbol and Einstein notation for the vector and scalar products \begin{align} |\vec v_2\times \vec v_3|^2&=\vec v_2\times \vec v_3\cdot\vec v_2\times \vec v_3=\epsilon_{ijk}v_{2j}v_{3k}\epsilon_{ilm}v_{2l}v_{3m}= \epsilon_{ijk}\epsilon_{ilm}v_{2j}v_{3k}v_{2l}v_{3m}\\ &=v_{2j}v_{3k}v_{2j}v_{3k}-v_{2j}v_{3k}v_{2k}v_{3j}=v_2^2v_3^2-(\vec v_2\cdot\vec v_3)^2\leq v_2^2v_3^3, \end{align} where I used the identity $\epsilon_{ijk}\epsilon_{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$, in which (the Kronecker delta) $\delta_{jl}$ is 1 if $j=l$ and 0 otherwise. Consequently, we obtain $$|\det A|\leq v_1v_2v_3,$$ where $v_i=|\vec v_i|$.

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