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Chapter 7 will focus on matrices $A$ with the property that $A^{T}=A$ . Exercises 23 and 24 show that every eigenvalue of such a matrix is necessarily real.Let $A$ be an $n \times n$ real matrix with the property that $A^{T}=A$ let $x$ be any vector in $\mathbb{C}^{n},$ and let $q=\overline{\mathbf{x}}^{T} A \mathbf{x}$ . The equalities below show that $q$ is a real number by verifying that $\overline{q}=q$ . Give a reason for each step.$$\overline{q}=\overline{\overline{\mathbf{x}}^{T} A \mathbf{x}}=\mathbf{x}^{T} \overline{A \mathbf{x}}=\mathbf{x}^{T} A \overline{\mathbf{x}}=\left(\mathbf{x}^{T} A \overline{\mathbf{x}}\right)^{T}=\overline{\mathbf{x}}^{T} A^{T} \mathbf{x}=q$$$$\begin{array}{llll}{\text { (a) }} & {\text { (b) }} & {\text { (c) }} & {\text { (d) }} & {\text { (e) }}\end{array}$$

a) $x ^ { T } = x ^ { T }$b) $\overline { A } = A ,$ because $A$ is renlc) $x ^ { T } A x$ is a scalar so transpose does not change itd) $( A B ) ^ { T } = B ^ { T } A ^ { T }$ - property of transpose operatore) $A ^ { T } = A$ and definition of $q$

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 5

Complex Eigenvalues

Vectors

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Harvey Mudd College

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in this video. We've been provided with the Matrix A and we're told it's in our and by in which is just a way to say it's a square matrix of size and by end, and all of its values are in the set of real numbers. Next, we're also told that a transpose equals a. This is a way of saying that the Matrix is also symmetric, and finally, we're going to pick a vector x from RN and define a new vector. Q. We see displayed below. This is a worry some way to define a vector because we have a vector times of matrix times a vector and that could well be undefined. So let's do a quick dementia analysis to see what this quantity even is. First. If X came from our end, then X transpose, whether we congregate it or not, is going to be of size one by N, since it will have now one row within entries. Now the Matrix A, which comes next. We're told already that it's of size and buy in, and so far so good Notice that these inner dimensions match and so this portion is well defined Next we multiply by a vector X so more and more interesting things continue to happen. And since X comes from our end, it's of size and buy one because it will now have en rose. But a single column for a column vector then noticed. This is still well defined because we next have in and one here for that matching dimensions. But one thing that's very interesting is the one and one here. It tells us that the result of Q equals X bar transpose times a X is an element in our so it's a real number or you could call it a scaler if you prefer. So let's see what happens with this particular scaler. Well, we said it's in our but we should be a little bit careful. Let's say a little bit more generically. It's a complex number in C. Then let's see if it really is a real number, which seems plausible. Let's calculate que transpose or excuse me que con? Trick it. By definition, we have to congregates the entire quantity here. So this is what it would look like First export transposed times a X, and were conjugating everything. And this expression Let's break this down step by step, where we know the first thing that could happen is if we consider the product here by association, we can rewrite the congregation as first take X contra git conjugate, but that gives us just X backs. Then we take the Conjure git of eight Times X, which would look like this open. And don't forget the transpose operator on X, so that's the set up so far. Next, let's break down the conjugal operator into what's inside. The parentheses will become X transpose. Then we'll have a congregate times ex con jacket. But a conjugal is a kind of a suspicious quantity because notice we said a came from our and by end, meaning all the numbers or all the entries in A are real numbers. And this means the congregate of a is just simply a so it's right that in on our next step, we now have X transpose. We have just the matrix A in place of ex con a con jacket. The ex itself can be a complex number, so we have to leave the congregate above it. Now where to go from here? We're trying to show that que con trick it is equal to queue so that it's a real number but also noticed that we can do the following The entry here we have is in C for a particular skit scaler. And that means since we're dealing with the scaler, we can transpose this entire thing and still get equality. Let's write that and break drop down a line and right X transpose times eight times x bar. Then again, a transpose here works because this is a scaler in si or ah, one by one matrix. Next thing we could do is use the transpose operator. It's going to transpose first This expert we see here and it becomes the first factor we take the transpose of a So there's a transpose here. Lastly, we take the transpose of X transpose, but that just gives us X back open. Don't forget the bar here that came from this first transposed. So this is very similar to when we took the conjure get of the conjure git on a first step the transpose of a transposed Just give this back X Now where to go from here? Notice. We've never used this assumption that a transpose is equal to a and that's suddenly very relevant here. Let's make that our next step. This is now equal to the congregate of X transpose than in place of a transpose. We can put a and then Times X, but notice. We've just matched this format here. So this is now equal to Q, which by our long chain of equality and by the transitive ity of Equality Q Bar or the Contra Give Q is equal to Q. And that implies that not only are we in C, but we can be more specific. Q is in our in other words, it's simply a real number.

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