Help with the question if the following is a linear transformation or not

Question

I am new to Linear Algebra, and am asked to check if the following transformation is linear. In principle, I know the criteria to check for: T(v+v')=T(v)+T(v') and c(Tv)=T(cv), where v is a vector, c is any scalar, and T is the transformation. However, when applying these to the following transformation, I get completely lost because of the complexity of the actual transformation. I am thinking there must be a shortcut I am not seeing. Any help would be very much appreciated.

$T:M^C_{nxn}\to M^C_{nxn}$ defined by $T(X)=2X+(1+3i)\bar X^t$. $M^C_{nxn}$ is a linear space over C.

There is an additional comment stating that we should refer to the general element of matrix $X$ as $x_{ij}$, while referring to the general element of matrix $\bar X$ as $\bar x_{ij}$

Thank you!

Answer 1

Hint:

Apply the rules mechanically:

$$2(X+X')+(1+3i)\overline{(X+X')}^t \\=2X+2X'+(1+3i)(\bar X+\bar X')^t \\=2X+2X'+(1+3i)(\bar X^t+\bar X'^t) \\=2X+2X'+(1+3i)\bar X+(1+3i)\bar X' \\=(2X+(1+3i)\bar X)+(2X'+(1+3i)\bar X').$$

There is nothing really complicated.

You may also just state that

• averaging is a linear operation,
• transposition is a linear operation,
• scalar multiplicationis a linear operation,
• addition is a linear operation.

Proving each of theses statements separately is at your reach.

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Update:

I didn't see that the overline denotes conjugation. Then

• complex conjugation is not is a linear operation.

Because $$\overline{aX}=\overline a\overline X\ne a\overline X.$$

Answer 2

The transformation clearly satisfies $T(X+Y) = T(X)+T(Y)$ right? Also if $\alpha$ is real then $T(\alpha X) = \alpha T(X)$ is trivial. However the final property we need for linearity is that $T(iX) = iT(X)$ for then we can conclude that $T((\alpha+i\beta)X) = \alpha T(X)+i\beta T(X)$ for real $\alpha,\beta$. We therefore consider

$$2iX+i(1+3i)\bar{X}^t=i(2X+(1+3i)\bar{X}^t) = iT(X) =^? T(iX) = 2iX-i(1+3i)\bar{X}^t$$

If we consider the identity matrix $I$ then is it true that

$$2iI+i(1+3i)I =^? 2iI-i(1+3i)I$$