Question

The trace of a square n×n matrix A=(aij) is the sum a11+a22+⋯+ann of the entries on its main diagonal. Let V be the vector space of all 2×2 matrices with real entries. Let H be the set of all 2×2 matrices with real entries that have trace 0. Is H a subspace of the vector space V?

Answer 1

Answer: Yes, H is a subspace of V

Step-by-step explanation:

We know that V is the space of all the 2x2 matrices with real entries.

H is the set of all 2x2 matrices with real entries that have trace equal to 0.

Obviusly the matrices that are in the space H also belong in the space V (because in H you have some selected matrices and in V you have all of them). The thing we need to prove is if H is an actual subspace.

Suppose we have two matrices that belong to H, A and B.

We must see that:

1) if A and B ∈ H, then (A + B)∈H

2) for a scalar number k, k*A ∈ H

lets write this as:

$A = \left[\begin{array}{ccc}a1&a2\\a3&a4\\\end{array}\right] B = \left[\begin{array}{ccc}b1&b2\\b3&b4\\\end{array}\right]$

where a1 + a4 = 0 = b1 + b4

then:

$A + B = \left[\begin{array}{ccc}a1 + b1&a2 + b2\\a3 + b3&a4 + b4\\\end{array}\right]$

the trace is:

a1 + b1 - (a4 + b4) = (a1 - a4) + (b1 - b4) = 0

then the trace is nule, and (A + B) ∈ H

and:

$kA = \left[\begin{array}{ccc}k*a1&k*a2\\k*a3&k*a4\end{array}\right]$

the trace is:

k*a1 - k*a4 = k(a1 - a4) = 0

so kA ∈ H

then H is a subspace of V

Answer 2

Let $\vec h$ and $\vec\eta$ be two vectors in $H$.

$H$ is a subspace of $V$ if (1) $\vec h+\vec\eta\in H$ and (2) for any scalar $k$, we have $k\vec h\in H$.

(1) True;

$\mathrm{tr}(\vec h+\vec\eta)=\mathrm{tr}(\vec h)+\mathrm{tr}(\vec eta)=0$

so $\vec h+\vec\eta\in H$.

(2) Also true, since

$\mathrm{tr}(k\vec h)=0k=k$

Therefore $H$ is a subspace of $V$.