Find the matrix of inner product space with orthonormal basis

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.

• 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. – Adrian Keister May 7 at 16:11
• 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, $A^TA=I_n$. – Shubham Johri May 7 at 16:11
• BTW, you may want to google "Gramian matrix" or "Gram Matrices" ... – 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: $$\;(4)\;|\det A|=1\;$$

• Isn't $\left|\begin{matrix}1 &0\\0 &-1\end{matrix}\right|=-1?$ – Adrian Keister May 7 at 16:16
• @AdrianKeister Yes....why? – DonAntonio May 7 at 16:18
• 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. – Adrian Keister May 7 at 16:53
• You can construct rotation matrices that are unitary, but are not equal to their own inverse, nor are they equal to their transpose. – Adrian Keister May 7 at 16:59
• 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. – DonAntonio May 7 at 21:05