Let $\mathbb{R}^n,\; n\geq 2$ be equipped with standard inner product. Let $\{v_1,v_2,......,v_n\}$ be $n$ column vectors forming an orthonormal basis of $\mathbb{R}^n$. Let $A$ be the $n\times n$ matrix formed by the column vectors $v_1,v_2,v_3,......,v_n$. Then

$1.$ $A = A^{-1}$

$2.$ $A = A^T$

$3.$ $A^{-1} = A^T$

$4.$ $\det(A) = 1.$

I have tried it as, since the given basis is orthonormal so the inner product $$\langle v_i,v_j\rangle = \begin{cases}0,\;& i\not=j\\ 1,&i=j \end{cases} ,$$

so we obtain the identity matrix i.e. $\det(A) = 1.$ But option $4$ is not correct, please someone help me to find the matrix.

  • $\begingroup$ If the columns are orthonormal, the matrix itself is unitary: $A^{-1}=A^{T}.$ The magnitude of its determinant is $1,$ but not necessarily the determinant itself. $\endgroup$ – Adrian Keister May 7 at 16:11
  • $\begingroup$ The $(i,j)^\text{th}$ element of $A^TA=[a_{ij}]_{n\times n}$ is the inner-product of the $i^\text{th}$ row of $A^T$ i.e. $v_i$, and the $j^\text{th}$ column of $A$ i.e. $v_j$;$$a_{ij}=\langle v_i,v_j\rangle=\begin{cases}|v_i|^2=1,&i=j\\0,&i\ne j\end{cases}$$Thus, $A^TA=I_n$. $\endgroup$ – Shubham Johri May 7 at 16:11
  • $\begingroup$ BTW, you may want to google "Gramian matrix" or "Gram Matrices" ... $\endgroup$ – DonAntonio May 7 at 16:17

As a huge hint, observe that if we write $A= (a_{ij})\;,\;\;i,j=1,...,n\;$ , then $\;A^t=(b_{ij})\;$ , with $\;b_{ij}=a_{ji}\;$ , and then

$$AA^t=(a_{ij})(b_{ij})=\left(\sum_{k=1}^na_{ik}b_{kj}\right)=\left(\sum_{k=1}^na_{ik}a_{jk}\right)=\langle v_i,\,v_j\rangle$$

where the last expression on the right denotes the usual, Euclidean inner product in $\;\Bbb R^n\;$ .

Work out this, using the fact that $\;\{v_1,...,v_n\}\;$ is an orthonormal basis, and get then almost at once (1)-(4) ...and in (4) you seem to have forgotten the absolute value: $\;(4)\;|\det A|=1\;$

  • $\begingroup$ Isn't $\left|\begin{matrix}1 &0\\0 &-1\end{matrix}\right|=-1?$ $\endgroup$ – Adrian Keister May 7 at 16:16
  • $\begingroup$ @AdrianKeister Yes....why? $\endgroup$ – DonAntonio May 7 at 16:18
  • $\begingroup$ So #4 is incorrect. For that matter, I don't buy 1 or 2, either. These are unitary matrices, so 3 is correct, but not necessary 1 or 2. $\endgroup$ – Adrian Keister May 7 at 16:53
  • $\begingroup$ You can construct rotation matrices that are unitary, but are not equal to their own inverse, nor are they equal to their transpose. $\endgroup$ – Adrian Keister May 7 at 16:59
  • $\begingroup$ I read, out of habit, $|\det A|=1\;$ , which is true for orthogonal or unitary matrices. 1-2 are correct as they are since $\;A^T=A^*\;$ in the real case, of course. $\endgroup$ – DonAntonio May 7 at 21:05

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