Write matrix as a linear combination of polynomials

Question

I can't figure out how to solve this:

Answer 1

For two vectors $n$-dimensional $x,y$ written as linear combinations of basis vectors $\{e_i\}$: $$x=\sum_{i=1}^n x_i e_i,\qquad y=\sum_{j=1}^n y_j e_j,$$ where $x_i$ and $y_i$ is the $i$'th component of $x$ and $y$, respectively, the linearity property of the inner product: $$(\alpha x+\beta y,z)=\alpha(x,z)+\beta(y,z)$$ for all vectors $x,y,z$ and all scalars $\alpha,\beta$, tells us that $$(x,y)=\left(\sum_{i=1}^n x_i e_i,\sum_{j=1}^n y_j e_j\right)=\sum_{i=1}^n\sum_{j=1}^n x_i y_j (e_i,e_j),$$ where $(e_i,e_j)$ is the Gram matrix $G_{ij}$:

$$G_{ij}=\begin{pmatrix} (e_1,e_1) & (e_1,e_2) & (e_1,e_3) & (e_1,e_4) \\ (e_2,e_1) & (e_2,e_2) & (e_2,e_3) & (e_2,e_4) \\ (e_3,e_1) & (e_3,e_2) & (e_3,e_3) & (e_3,e_4) \\ (e_4,e_1) & (e_4,e_2) & (e_4,e_3) & (e_4,e_4) \end{pmatrix}.$$

If $G_{ij}=\delta_{ij}$ (see kronecker delta) i.e. if $G=I$, that is if the basis set $\{e_i\}$ is orthonormal, this produces the ordinary formula for the inner product: $$(x,y)=\sum_{i=1}^n x_iy_i=x_1y_1+x_2y_2+\dots+x_ny_n,$$ because $\sum_{j=1}^n y_j \delta_{ij}=y_i$, which is the same as saying $Iy=y$ just expressed in index notation.

So since we know that $$(x,y)=\sum_{i=1}^n\sum_{j=1}^n x_i y_j G_{ij}$$ we can write that $$(x,x)=\sum_{i=1}^n\sum_{j=1}^n x_i x_j G_{ij}=\left(\sum_{i=1}^n x_i G_{ij}\right)\left(\sum_{j=1}^n x_j G_{ij}\right),$$ because of a property of double sums (see double series)

So $G_{2j}$ for instance is the 2nd row of $G$ and $G_{i3}$ is the 3rd column of $G$. This gives us

$$(w,w)=(xG_{1j}+yG_{2j}+zG_{3j}+tG_{4j})(xG_{i1}+yG_{i2}+zG_{i3}+tG_{i4})=\big[(2x,x,-2x,-x)+(y,y,2y,y)+(-2z,2z,5z,z)+(-t,t,t,2t)\big]$$ $$\left[\begin{pmatrix} 2x\\ x\\ -2x\\ -x \end{pmatrix}+\begin{pmatrix} y\\ y\\ 2y\\ y \end{pmatrix}+\begin{pmatrix} -2z\\ 2z\\ 5z\\ z \end{pmatrix}+\begin{pmatrix} -t\\ t\\ t\\ 2t \end{pmatrix}\right]$$ Now we have the product of the sums of 4 row vectors with the sums of 4 column vectors and also we see that these are actually the same vectors just in row and column form respectively this is because $G_{ij}=G_{ji}$ that is $G$ is symmetric which is a consequence of the fact that the (real) inner product is symmetric i.e. $(x,y)=(y,x)$.

Observe that $$(x_1,x_2,x_3,x_4)\begin{pmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{pmatrix}=x_1^2+x_2^2+x_3^2+x_4^2.$$ This tells us that all we have to do is sum up the 4 row vectors and then take the sum of the squares of the components to find $(w,w)$: $$(w,w)=(2x+y-2z-t)^2+(x+y+2z+t)^2+(-2x+2y+5z+t)^2+(-x+y+z+2t)^2.$$

Does this make sense? Please write otherwise. That's it for the first question now see if you can figure out the second one yourself :)