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[Extra Credit] Prove the following statement: If T € L(V) and dim(T(V)) 1, then T2 aT for some a € R (in other words, for all v € V we have T2 (v)...

Question

[Extra Credit] Prove the following statement: If T € L(V) and dim(T(V)) 1, then T2 aT for some a € R (in other words, for all v € V we have T2 (v) aT(v) for some fixed real number a)

[Extra Credit] Prove the following statement: If T € L(V) and dim(T(V)) 1, then T2 aT for some a € R (in other words, for all v € V we have T2 (v) aT(v) for some fixed real number a)



Answers

If $a, b, c, d$ are real numbers and $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, prove that $A^{2}-(a+d) A+(a d-b c) I=O$.

Really given problem we want to show that a uh the absolute value of a minus the absolute value D. Is less than or equal to the absolute value of a minus T. So um one way that we can express this, we can follow the hint or we can just think through this. So if we have the absolute value of uh the absolute value of A. Within the absolute value we have a minus the absolute value of E. Yes. Mhm. We want to show that's less than or equal to. Yes the value of A minus B. Well considering our different scenarios we see that in this case there will always be a subtraction taking place. This can never become a minus and negative so we can never add these two together. We're always going to be subtracting them together and then we'll be taking their absolute value in the same way we'll be taking the absolute value here. So really um what we see is that A minus B is always going to be um could be a subtraction or it could be an addition. But regardless when we take its absolute value we're turning it positive. However here it's always going to be a subtraction or turning it positive so this one will always be greater than two.

We're going to do problem number 14. Okay. Let me just, uh, did once the question on the question is, keep and complete the proof of this statement for all riel numbers. Okay, for proving a plus B Manus be Is it calls to a plus B plus Manus be okay. So let me write the first statement which is given to us. That is a plus. B. Manus be okay. This is Sheridan s, A plus B on the plus. Manus, be okay. The reason is treated there. That is definition offs. Obstruction. Okay, definition off subtraction. No. Let us proceed towards the second step. The second step is it is written as it is. That is a plus. B Manus B is equals toe. It is written as it is. Plus, it had taken be on plus minus B at bracket in back it okay, and it is written the property here. That associative associative property offs. Obstruction. Now let's just do the third state that third It is there for this. That is a plus. B minus. B is written as a plus zero. So what we can say? What is the reason here? What was the reason. If we see it is Donna a plus, uh, be managed. Be ok. A plus Be managed. Be so help B and B both are same. So we can say that that is inverse off subtraction. Okay, that enforce offs obstruction. So you can write that Also, you can simply right substructure in. Okay. On what property it follows. It follows in horse this month. We have to write now it is written that a plus B manage b is equals toe A So what we can say if we at zero to certain number. So that is the property we call it Identity property. Okay, Identity property. So these are the reasons which you will right, while doing this and complete the solution, That's all. Thank you.

Yeah, okay. Our job is to prove that the absolute value of b minus a is equal to the absolute value of a minus B for all and be so yeah, one thing we know if B minus A is negative then a minus B is positive. So let's say b minus A is negative if the mine is A is negative and the absolute value of b minus A. And this implies that a minus B is positive. If the minus A is negative then the absolute value of B minus A is equal to the opposite B minus A, which is negative B minus A which is equal to a minus B. And since a minus B is greater than zero, the absolute value of a minus B is equal to a minus B and a minus B equals a minus B, which is half of what we were supposed to prove. Case two A -7 is negative, which implies b minus is positive. So the absolute value of a minus B is equal to the opposite of a minus B or b minus A. The absolute value of b minus A is equal to b minus A, and b minus A is equal to b minus A. And then case three, the trivial case A minus B equals zero, and that implies B minus A equals zero and zero equals zero, So there you go.

Hello everyone. The time average of the function after function is to find us His capital one upon teeth integration of function jude S. T. For the limited minus key way to keep this T. V. Two average value of sinus square of omega T. one way to 1- sign up omega T upon omega T into cause off. Who are you? No has been assigned a square theater plus causes square theater is cut to one. So using Sandy Square the topless causes square theaters called to one. The average value of cross the square of omega T Build Me 1- Average my work Sinus square omega T. Substituting the value. You will get one minus half one minus sine of omega T upon omega T. Cause of two negative. So finally you will get the answer that is half one plus sign off omega t upon omega T. Cause of to omega T. That so thanks for watching it.


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