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Let $A$ be a real $n \times n$ matrix, and let $x$ be a vector in $\mathbb{C}^{n} .$ Show that $\operatorname{Re}(A \mathbf{x})=A(\operatorname{Re} \mathbf{x})$ and $\operatorname{Im}(A \mathbf{x})=A(\operatorname{Im} \mathbf{x})$

See the proof.

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 5

Complex Eigenvalues

Vectors

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in this video, we're starting off with the Matrix A which is going to be square of size and by n and contain rial number entries. Then we're gonna pick a vector X that comes from the space off complex numbers within entries as a column vector. We're going to prove that the real part of eight Times X is equal to a times a real part of X and then likewise for the imaginary part. So to start out with this kind of proof, let's do the following. I'm going to use the fact that X came from CNN to say Let X be equal to U plus I times V where you and V have only real number entries or, in other words, U and V are in our end, So any vector X from CNN can be decomposed in this way. Now that we have this decomposition, let's also list that the real part of X is then equal to you and the imaginary part of X is then equal to V. So now let's start verifying these equations. Let's work with this one here. We're going to verify that the real part of a Times X is eight times the real part of X, so let's consider a Times X first eight times X by substitution is eight times the vector U plus ivy so we can make a substitution here and then matrix multiplication distributes across a vector some so we'll have a times you, plus I times a times B. Then we can state the following the real part of a Times X is this quantity here. So the real part of eight Times X is a times you, but the same time we know that you is equal to the real part of X as well. So this is now equal to eight times the real part of X, and that verifies this first equation. Let's do this as well. For the imaginary part we have that the imaginary part of a X recall. We compute x here, and this is the imaginary portion of that computation. So it's equal to eight times V. And then, as before, we know that V is the imaginary part of X. Let's make that substitution as well. We have a times the imaginary part of X, and that's precisely what we want to show here. So now are proof is complete

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