Answer:
H is not a subspace of v
Step-by-step explanation:
Please see attachment
Answer:
tr(zA+B) = za+zb + a1+b1 = z(a+b)+(a1+b1) = 0+0 = 0 .
Step-by-step explanation:
You need to show that given two matrices A,B such that tr(A) = tr(B) = 0 and a random number "z", tr(zA+B) = 0.
[tex]A = \left[\begin{array}{cc}a&x\\y&b\end{array}\right] \\B = \left[\begin{array}{cc}a1&x1\\y1&b1\end{array}\right] \\[/tex]
By hypothesis a+b=0 and a1+b1 = 0
tr(zA+B) = za+zb + a1+b1 = z(a+b)+(a1+b1) = 0+0 = 0
Therefore H would be a subspace of V.
