In $\mathbb{R}^3$, we have the basis $B=\left\{\vec{b_1};\vec{b_2};\vec{b_3}\right\}$ with $\vec{b_1}=\begin{pmatrix} 2\\ 2\\ 3 \end{pmatrix}, \vec{b_2}= \begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}, \vec{b_3}=\begin{pmatrix} 2\\ 1\\ 1 \end{pmatrix}$ and in terms of this basis, we have a scalar product by the Gramian matrix $A=\begin{pmatrix} 4 & 1 & 2\\ 1 & 3 & 2\\ 2 & 2 & 5 \end{pmatrix}$. Determine the Gramian matrix in terms of the standard basis $B^3_0$.

I'm not sure how to do this and I couldn't find much information about Gramian matrix on the internet either. So first of, by standard basis $B^3_0$ it's probably meant that you use Gaussian elimination, and that we use the basis together with the unit matrix, so then $B^3_0$ must be simply the basis in form of a matrix, $B^3_0 = \begin{pmatrix} 2 & 1 & 2\\ 2 & 1 & 1\\ 3 & 1 & 1 \end{pmatrix}$.

The gramian matrix is defined as a set of vectors in an inner product space, so I guess in this case these vectors are the columns of $A$.

And now to get the gramian matrix of $A$, do I just need to multply $A$ with its transpose? On the other hand, this shouldn't be the correct solution because it must be in terms of $B^3_0$ and we woulddn't have made any use of it..? : /


1 Answer 1


In general, given an $n$-dimensional linear space $V$ endowed with inner product $(\alpha, \beta)$, suppose the Gramian matrices of $(\alpha, \beta)$ under two different bases $\{\alpha_1, \ldots, \alpha_n\}$ and $\{\beta_1, \ldots, \beta_n\}$ are $G_1$ and $G_2$, and $$(\beta_1, \ldots, \beta_n) = (\alpha_1, \ldots, \alpha_n)P \tag{1}$$ for some non-singular matrix $P$, then $G_2 = P'G_1P$. This result can be restated as that the Gramian matrices for an inner product under two different bases are congruent to each other.

In your case, take $\beta_i = b_i, \alpha_i = e_i, i = 1, 2, 3.$ It is easily seen that the transition matrix (as $(\alpha_1, \alpha_2, \alpha_3) = I_{(3)}$, the $3 \times 3$ identity matrix.) $$P = B = \begin{bmatrix} 2 & 1 & 2 \\ 2 & 1 & 1 \\ 3 & 1 & 1 \end{bmatrix}.$$

In addition, $G_2 = A$, therefore by $(1)$ the Gramian matrix under $\{\alpha_1, \alpha_2, \alpha_3\}$ is given by $$G_1 = (P')^{-1}AP^{-1}.$$

It should not be difficult to calculate the inversion of $P$ and the matrix multiplication from here.

  • $\begingroup$ Thank you for answer! I have just one question, what is the matrix $P'$? $\endgroup$
    – eyesima
    Jun 2, 2018 at 23:45
  • $\begingroup$ The transpose of $P$. $\endgroup$
    – Zhanxiong
    Jun 3, 2018 at 0:03

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