Let A be an n×n n times n real matrix with the property that AT=A. eh to the t , equals eh . Show that if Ax=λx eh x equals lamda x for some nonzero vector x in Cn, blackboard bold cap c to the n , comma then, in fact, λ lamda is real and the real part of x is an eigenvector of A. [Hint: Compute x¯¯TAx, x bar to the t eh x comma and use Exercise 23. Also, examine the real and imaginary parts of A x.]

The statement is sometimes true.

Step-by-step explanation:

The statement is sometimes true, ie when x is positive integer, then -3x will be negative.

But when x is negative, then -3x will be positive { -3 * - x = +3x }

If Ax-2x for some scalar λ, then x is an eigenvector of A because the only solution to this equation is the trivial solution.

Step-by-step explanation:

The eigen vector is linear transformation which is a non zero vector which changes due to scalar factor. The eigen factor value is denoted by lambda. The eigen vector is a latent vector.

Sometimes true! Nonzero can be all positive and all negative! -3(-3)=9 this is an example of why it’s false|| -3(4)=-12 is an example if why it’s true||

c

Step-by-step explanation:

1/x^4 is your answer

sometimes true

Step-by-step explanation:

if x is a positive integer the statement is true, if x is negative integer the statement will be false.

I am pretty sure it is always true

The answer to your question is C

Step-by-step explanation:

Resulting polynomial contains a maximum mn positive terms.

Step-by-step explanation:

Given one polynomial contains m nonzero terms and second polynomial contains n nonzero terms.

To show after multiplication and combining similar terms how many positive terms contan by both polynomial.

Now let,

m=1=(a), n=2=(x,y) then after multiplication a(x+y)=ax+ay, 2 positive terms.m=2=(a,b), n=3=(x,y,z) then after multiplication (a,b)(x+y+z)=ax+ay+az+bx+by+bz, 6 positive terms.m=3=(a,b,c), n=4=(x,y,z,t) then after multiplication (a+b+c)(x+y+z)=ax+ay+az+at+bx+by+bz+bt+cx+cy+cz+ct, 12 positive terms.

So we see that after multiplication of m and n positive terms, there are mn positive terms are there.

To prove this we have to apply mathematical induction. So let the statement is true for m=p and n=q number of positive terms, then mn=pq.

We have to show avobe ststement is hold for m+1, n+1. Considering,

(m+1)(n+1)=mn+m+n+1=pq+p+q+1=p(q+1)+1(q+1)=(p+1)(q+1)

Hence above statement is true for m+1 and n+1.

Thus there will be mn nonzero terms after multiplication and combine positive terms.

Sometimes

Step-by-step explanation:

When x is a nonzero integer, then x can be

negative;positive.

If x is negative, then [tex]-3x[/tex] is positive (multiplying negative number by -3 you get positive number).

If x is positive, then [tex]-3x[/tex] is negative (multiplying positive number by -3 you get negative number).

So, the statement "When x is a nonzero integer, the quantity -3x will be negative" is sometimes true.