# Finding tensor components via matrix manipulations

## Homework Statement

Imagine we have a tensor ##X^{\mu\nu}## and a vector ##V^{\mu}##, with components

##
X^{\mu\nu}=\left( \begin{array}{cccc}
2 & 0 & 1 & -1 \\
-1 & 0 & 3 & 2 \\
-1 & 1 & 0 & 0 \\
-2 & 1 & 1 & -2 \end{array} \right), \qquad V^{\mu} = (-1,2,0,-2).
##

Find the components of:

(a) ##{X^{\mu}}_{\nu}##
(b) ##{X_{\mu}}^{\nu}##
(c) ##X^{(\mu\nu)}##
(d) ##X_{[\mu\nu]}##
(e) ##{X^{\lambda}}_{\lambda}##
(f) ##V^{\mu}V_{\mu}##
(g) ##V_{\mu}X^{\mu\nu}##

## The Attempt at a Solution

(a) ##{X^{\mu}}_{\nu}=X^{\mu\rho}\eta_{\rho\nu}=\left( \begin{array}{cccc}
2 & 0 & 1 & -1 \\
-1 & 0 & 3 & 2 \\
-1 & 1 & 0 & 0 \\
-2 & 1 & 1 & -2 \end{array} \right)
\left( \begin{array}{cccc}
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{array} \right)=\left( \begin{array}{cccc}
-2 & 0 & 1 & -1 \\
1 & 0 & 3 & 2 \\
1 & 1 & 0 & 0 \\
2 & 1 & 1 & -2 \end{array} \right)
##,

where the rows of the left matrix are multiplied by the columns of the right matrix because the summation is over the second index of ##X^{\mu\rho}## and the first index of ##\eta_{\rho\nu}##.

(b) ##{X_{\mu}}^{\nu}=\eta_{\mu\rho}X^{\rho\nu}=
\left( \begin{array}{cccc}
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{array} \right)
\left( \begin{array}{cccc}
2 & 0 & 1 & -1 \\
-1 & 0 & 3 & 2 \\
-1 & 1 & 0 & 0 \\
-2 & 1 & 1 & -2 \end{array} \right)
=\left( \begin{array}{cccc}
-2 & 0 & -1 & 1 \\
-1 & 0 & 3 & 2 \\
-1 & 1 & 0 & 0 \\
-2 & 1 & 1 & -2 \end{array} \right)
##,

where the rows of the left matrix are multiplied by the columns of the right matrix because the summation is over the second index of ##\eta_{\mu\rho}## and the first index of ##X^{\rho\nu}##.

(c) ##X^{(\mu\nu)}=\frac{1}{2}(X^{\mu\nu}+X^{\nu\mu})=\frac{1}{2}\Bigg[\left( \begin{array}{cccc}
2 & 0 & 1 & -1 \\
-1 & 0 & 3 & 2 \\
-1 & 1 & 0 & 0 \\
-2 & 1 & 1 & -2 \end{array} \right)+\left( \begin{array}{cccc}
2 & -1 & -1 & -2 \\
0 & 0 & 1 & 1 \\
1 & 3 & 0 & 1 \\
-1 & 2 & 0 & -2 \end{array} \right)
\Bigg]=\left( \begin{array}{cccc}
2 & -0.5 & 0 & -1.5 \\
-0.5 & 0 & 2 & 1.5 \\
0 & 2 & 0 & 0.5 \\
-1.5 & 1.5 & 0.5 & -2 \end{array} \right)
##

(d) ##X_{[\mu\nu]}=\frac{1}{2}(X_{\mu\nu}-X_{\nu\mu})=\frac{1}{2}(\eta_{\mu\rho}X^{\rho\sigma}\eta_{\sigma\nu}-\eta_{\nu\sigma}X^{\sigma\rho}\eta_{\rho\mu})##

Are my answers to (a), (b) and (c) correct?

With part (d), I'm not sure if I should take the original matrix to ##X^{\rho\sigma}## or the transposed matrix to ##X^{\rho\sigma}##? Does it make a difference anyway?

andrewkirk
Homework Helper
Gold Member
What you have written looks correct to me, including (d). If you instead used the transposed matrix in (d) you would change just the sign of the answer.

Ok!

(d) ##X_{[\mu\nu]}=\frac{1}{2}(X_{\mu\nu}-X_{\nu\mu})=\frac{1}{2}(\eta_{\mu\rho}X^{\rho\sigma}\eta_{\sigma\nu}-\eta_{\nu\sigma}X^{\sigma\rho}\eta_{\rho\mu})=
\frac{1}{2}\left( \begin{array}{cccc}
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{array} \right)
\left( \begin{array}{cccc}
2 & 0 & 1 & -1 \\
-1 & 0 & 3 & 2 \\
-1 & 1 & 0 & 0 \\
-2 & 1 & 1 & -2 \end{array} \right)
\left( \begin{array}{cccc}
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{array} \right) -
\frac{1}{2}\left( \begin{array}{cccc}
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{array} \right)
\left( \begin{array}{cccc}
2 & -1 & -1 & -2 \\
0 & 0 & 1 & 1 \\
1 & 3 & 0 & 1 \\
-1 & 2 & 0 & -2 \end{array} \right)
\left( \begin{array}{cccc}
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{array} \right)##
##=\frac{1}{2}
\left( \begin{array}{cccc}
2 & 0 & -1 & 1 \\
1 & 0 & 3 & 2 \\
1 & 1 & 0 & 0 \\
2 & 1 & 1 & -2 \end{array} \right)-\frac{1}{2}
\left( \begin{array}{cccc}
2 & 1 & 1 & 2 \\
0 & 0 & 1 & 1 \\
-1 & 3 & 0 & 1 \\
1 & 2 & 0 & -2 \end{array} \right)
=\left( \begin{array}{cccc}
0 & -0.5 & -1 & -0.5 \\
0.5 & 0 & 1 & 0.5 \\
1 & -1 & 0 & -0.5 \\
0.5 & -0.5 & 0.5 & 0 \end{array} \right)
##

(e) ##{X^{\lambda}}_{\lambda} =\eta_{\lambda\rho}X^{\rho\sigma}\eta_{\sigma\lambda}##

Is my answer to (d) correct?

Am I on the right track with (e)? How do I sum over ##\lambda##?

andrewkirk
Homework Helper
Gold Member
(d) looks OK
(e) is just the trace of a matrix you have already calculated (where?), so you don't need to do any new matrix multiplications.

(d) looks OK

Thanks!

(e) is just the trace of a matrix you have already calculated (where?), so you don't need to do any new matrix multiplications.

(e) ##{X^{\lambda}}_{\lambda}={X^1}_{1}+{X^2}_{2}+{X^3}_{3}+{X^4}_{4}=-2+0+0-2=-4##, from part (a).

(f) ##V^{\mu}V_{\mu}=
\left( \begin{array}{cccc}
-1 & 2 & 0 & -2 \end{array} \right)
\left( \begin{array}{c}
-1 \\
2 \\
0 \\
-2 \end{array} \right)=9
##

(g) ##V^{\mu}X^{\mu\nu}=
\left( \begin{array}{cccc}
-1 & 2 & 0 & -2 \end{array} \right)
\left( \begin{array}{cccc}
2 & 0 & 1 & -1 \\
-1 & 0 & 3 & 2 \\
-1 & 1 & 0 & 0 \\
-2 & 1 & 1 & -2 \end{array} \right)=
\left( \begin{array}{cccc}
0 & -2 & 3 & 9 \end{array} \right)
##

What do you think?

$V_{\mu}$ and $V^{\mu}$ have different components - don't forget that you need to apply the metric tensor to raise and lower indices!

$V_{\mu}$ and $V^{\mu}$ have different components - don't forget that you need to apply the metric tensor to raise and lower indices!

Ok!

(f) ##V_{\nu}=V^{\rho}\eta_{\rho\nu}=
\left( \begin{array}{cccc}
-1 & 2 & 0 & -2 \end{array} \right)
\left( \begin{array}{cccc}
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{array} \right)
=
\left( \begin{array}{cccc}
1 & 2 & 0 & -2 \end{array} \right)
##

Therefore, ##V^{\mu}V_{\mu}=V^{0}V_{0}+V^{1}V_{1}+V^{2}V_{2}+V^{3}V_{3}=(-1)(1)+(2)(2)+(0)(0)+(-2)(-2)=7##.

Is it correct now?

Yup, looks correct now. Same for part (g) - based on your original post, it seems to be $V_{\mu}$ instead of $V^{\mu}$. It helps to remember that in the Einstein summation convention, one index should be superscripted and the other subscripted.

Yup, looks correct now. Same for part (g) - based on your original post, it seems to be $V_{\mu}$ instead of $V^{\mu}$. It helps to remember that in the Einstein summation convention, one index should be superscripted and the other subscripted.

Ok, so using ##V_{\mu}## from part (f),

(g) ##V_{\mu}X^{\mu\nu}=
\left( \begin{array}{cccc} 1 & 2 & 0 & -2 \end{array} \right)
\left( \begin{array}{cccc}
2 & 0 & 1 & -1 \\
-1 & 0 & 3 & 2 \\
-1 & 1 & 0 & 0 \\
-2 & 1 & 1 & -2 \end{array} \right)= \left( \begin{array}{cccc} 4 & -2 & 5 & 7 \end{array} \right).##

Is it all right?

Yup, seems alright to me.

Thanks to both andrewkirk and Fightfish for helping me to solve the problem!