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Orthogonally diagonalize the matrices in Exercises $13-22,$ giving an orthogonal matrix $P$ and a diagonal matrix $D .$ To save you time, the eigenvalues in Exercises $17-22$ are: $(17)-4,4,7 ;(18)$ $-3,-6,9 ;(19)-2,7 ;(20)-3,15 ;(21) 1,5,9 ;(22) 3,5$
$$
\left[\begin{array}{rrr}{5} & {8} & {-4} \\ {8} & {5} & {-4} \\ {-4} & {-4} & {-1}\end{array}\right]
$$

Were given a matrix A and were asked toward fog only diagonal eyes. This matrix giving an orthogonal matrix diagram matrix were also given that the Eigen values for this Matrix or Lambda equals three and five. The Matrix A is 401004101104001 04 Noticed that a transposed says the Matrix has the same diagonal for four. For four. We exchanged zero with zero one with one zero with zero zero with zero one with one and zero with zero and recognize that this is the same as the matrix. A. So it follows. That's since a transfers equals a is a symmetric matrix. Therefore, it follows that a is north Ogden Aly Diagonal Izabal Your thought only diagonal eyes A when he defined complete set of orthe normal vectors. Eigen vectors for a So take lambda One to be the first Eigen value three. Then we have that. If the one is an Eigen vector then satisfies the equation. A minus three i v one equals zero vector The matrix a minus three I is the same as matrix except for before has become ones So we obtained the matrix. 10100101 10100101 in this simply reduces to the matrix. 10100101 0000 0000 from which we obtained a system of equations V 11 plus V one of three equals zero and V 12 plus view and four equals zero. This tells us that the 13 equals negative view on one and the 14 equals negative b 12 And so if we take V 11 to be the parameter s and he one to be the parameter t, we get that The general form for the I conductor V one is the vector s T negative s negative T which would be written as a linear combination s times 010 negative one or zero 01 My mistake s times 10 negative 10 plus t times +010 Negative one. So we see that this wagon spaces two dimensional. And if we take s to be one and t v zero, we obtained one Aiken Victor called V one is the Agon vector 10 negative one zero it would take Has to be zero. On TV one, we obtain a second Eigen Vector V two, which is linearly independent from V one, which is the connector 01 zero negative one. And so we have that set V one be to is a basis for this Eigen space. They also have that V one dotted with feet too is equal to zero plus zero plus zero plus zero, which is zero. And so we have that V one is or thought Donald to V two. And therefore we have this set view and be two of the north Ogle set norm of V one is the square root of one plus one, which is Route two in the normal. V two is the square root of one plus one, which is Route two. And so we have the unit vectors. You one, which is V one over. Norm of the one is equal to 1/2 zero negative. One of a route to zero. And the unit after you too, is the vector V two over the normal vector he to which is 0 1/2 zero negative 1/2. And we have that both. You want you to our Eigen vectors and are linearly independent in North Ogle. So that set you on you too. Isn't Ortho normal basis for this Eigen space? Next, Let claimed to be the second and last Eigen value five. Then if the three is an Eigen vector associated with this Eigen value, we have that A minus five I times feet three is equal to the zero vector matrix. A minus five. I is the sings matrix A Except for the diagonals now negative one. Instead of positive one, we obtain the matrix. Negative. 1010 ones. Zero negative. One 01 10 Negative. 10 and 010 Negative one. This could be reduced to the matrix with reduction 10 Negative. 10 010 Negative. One and 00000000 from which we can obtain these system of equations. The 31 minus V 33 equals zero and be three to minus fee. 34 equals zero. So we have that. These 31 is equal to be 33 and V 32 is equal 33 4 So if we take B 31 to be the parameter s and V 32 to be the parameter t yet the general form for the I conductor V three is S T yes, T is equal to the linear combination s times 1010 plus T times +01 zero one And if we take us to be won t v zero weekends the Eigen Vector V three, which is 10 10 And if we take s to be zero and t to be one, get second Eigen Victor, before which is 0101 and by construction we have that these two wagon vectors are linearly independent and thus form a basis for this wagon space. We also have that V three died with before is equal to zero, which implies that V three is orthogonal to before. So it's actually in orthogonal basis for this vector space. The norm of V three is equal to the square root of one plus one which is to in the normal before is the square root of one plus one which is equal to. And so we have the unit Vector you three, which is equal to V three over the norm of you three, Our normal V three. My mistake is a vector one over route to zero 1/2 zero. And the vector You four is a unit vector, which is V four over the norm of the four. This is the vector zero. Whatever too zero one of the route to and we have that both u three and U four our Eigen vectors for this wagon space and are valuable and are unit vectors. So it follows that set you three. You four is an Ortho Normal aces for this Eigen space. Since our Matrix A is symmetric, it follows that you won and you two are Ortho or thought Donald U three and U four. So it follows that to set u one u two you three you for is a complete Ortho normal set of Eigen vectors for this matrix. And we have that. If we take P to be the Matrix who's calling Victor's Air U one U two, you three and you four. This is the Matrix one of a right to zero negative one of a route to 01 of a route to zero negative 1/2 and whatever to zero Whatever. Route to zero and zero one of a route to 0 1/2 when we take matrix D to be a matrix whose entries our gagan values on the diagonal. So you have each Agnelli was multiplicity too. And so this is the matrix 3355 and then zeros at the worlds. And using these two matrices p and G, we can diet.

Three miners well minus two for two. We know that. So they exist. Three alien values. Oh, God. Lambert One equals two minus land to it too. And lamb that he also seven? No. When Lambda One equal to minus. Then I convicted. We want equality minus one one by two. Like what? And when the again Really the seven. Then I can Victor minus one by two. One judo. And when I die again, really, seven. When I get there is also minus one way to one. No done. Normally taken Victor's, uh, you won equals minus two by three minus one by three. And who by and taken? Not much. I can make that use you. Minus one day you fight, Who made a fake? You go and you three equals It is also minus one way to fight the baby's fine deal and let be equals minus two by three one lately go minus on Little fight. We'll wait. Right deal and remains to fight you and they're viable. Metric dee Minus Who? Jiggle Jiggle deedle seven judo deedle deedle Thank you for watching

So they want us to or thought analyzes. And since they already give us the Eigen vectors arming the Eigen values, we could just go ahead and find the Eigen vectors and then normalize them and then plug everything in from there. So let's first do for when Lambda is equal to negative three. So we do a minus minus three. I. So that's really just like we're gonna add three along the diagonal. So this is going to be for negative 64 negative 65 negative to four negative 20 And if we go ahead and roll reduces, this will end up giving us one zero negative one half 01 negative one and then all zeros down here. And so what this implies is that x one minus one half x three is equal to zero. So that's just x one is even told one half x three And then over here X two minus x 30 or X two is a good X three so we could go ahead and write. So we have x one next two x three here, so x one we said Waas one half x three x two is just x three and the next three is X three. So we go ahead and pull that out. So it be x 31 half 11 Now, before we actually normalize this, I don't have toe square fraction. So I'm just gonna multiply this by two. So it won't really change anything since when we normalize it. That'll get accounted for. So just be 1 to 2. So now let's normalize this one. So to do that, it would be one squared plus two squared plus two squared spur voted. So that would be, um Well, what I do to here, Right that to square. So that would be four plus 48 plus 19 So that's actually just gonna be three. So we divide all of this by three. So that actually gives us when Lambda One is even to negative three. Our first Eigen vector is going to be one third, two thirds and two thirds. All right, now we'll do the same thing. But with Lambda equal to negative six. Um, let me pick this up and scoop this down. Just we kind of have it with us. Eso would be a minus minus six eyes. So again, it's gonna be like we just add six to all of this. So this is going to be equal to s 0785 and everything else stays the same. And now, if we were to go ahead and roll reduce this, we should end up with. Actually, this is in a five. Here. Um, there's maybe three plus three are negative. Three plus six is three, right, but only very reduced. Now, this one that should give us 101 are 011 half and then 000 So then this implies that x one plus x three is equal to zero. Um, which is gonna be next one is equal to negative x three and then for the next one. This is saying x two plus one half x three is going to be zero or X to negative one half x three so that we can use that to build our Eigen vector. So we have x one x two x three So x one was negative x three x two is negative one half x three and x three is just x three So this is x three negative one negative one, House one. And just like before, I'm gonna multiply all this by two. Uh, just cause I don't wanna have toe square a fraction. So this is going to be negative, too. Negative one to. So now we can normalize this one on doing that would get negative two squared plus negative one squared plus two squared, all square rooted. So that would be four plus 48 plus one. So, again, three once described it. So then that tells us when lambda, uh, two is equal to negative six. We get our second director being so we just divide all that by three. So negative two thirds negative. One third and to third. And then our last one. We're going to do this with nine. Let's go ahead and scoot this town and going to now do a minus nine I. So we just tracked nine along the diagnosis. That be negative. Eight negative seven and negative 12. And then everything else stays the same. And then when we rode use this, we should end up getting eso. It's 10 negative too. 012 and then 000 So then this implies that x one minus two x three is equal to zero or x one is equal to x three. And in that second equation is saying. X two is x two plus two x three is equal to zero or x two is negative two x three and then we go ahead and plug that into x one x two x three. So then again, X one is, uh, it's just a equal sign, uh, two x three x two is negative two x three and x three is just x three. Pull that out. So we have to negative 21 and then we'll go ahead and normalize this one. And doing that, we should get it's gonna be two squared plus negative two squared plus one square rooted. So again, that's just going to be, uh, Route nine or three. So then that would give us when Lambda is equal to nine. Uh, our third Eigen vector is going to be too negative to one, or we need to divide all these by three first, so it would be two thirds negative two thirds and one third. So let's go ahead and write out with our diagonal ization matrix system. This is gonna be D is equal to, um So we had negative six negative. Three and nine. Remember? Doesn't remember the order you write these a song is when you write your or thought little matrix. You kind of line everything up. But I'm just used to draw writing it from smallest to largest like this. All right. And so now let's write what are vectors are actually the one associate with nine here, So if we write, p is the one we just found. So this is that two thirds negative. Two thirds. One third now for negative three. Uh, this should be actually, let me just pick these up and scoop them all down. Then there will be easier for me to just plug it. So he's down. Okay, so now for negative three, we have of one third, two thirds to third and then or negative six, we have negative two thirds negative. One third to third. And if you want, you could go ahead and pull out that one third just to make it look a little bit prettier. So this isn't really needed, but it will make it look better s that we have negative too negative. 12 1 to 2. And then to negative. To what? So now these are going to be our diagonal ization, along with far orthogonal matrix P.


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