let-u-be-an-n-times-n-orthogonal-matrix-show-that-the-rows-of-u-form-an-orthonormal-basis-of-mathbbr

Question

Let $U$ be an $n 	imes n$ orthogonal matrix. Show that the rows of $U$ form an orthonormal basis of $\mathbb{R}^{n}$ .

Foster Wisusik · Accepted Answer

Okay, This question gives us an orthogonal matrix, and it wants us to prove that the columns of this matrix form an Ortho normal basis for our end. So we&#x27;ll start by writing you as this matrix, which is spanned by these vectors u one u two all the way up to U N. And if we&#x27;re just asked to prove that these vectors form a north a normal basis, we need to prove that they form a basis and that they have magnitude one. And there are Huggel. So first we know that based on definition oven orthogonal matrix you I dotted with you. Jay is equal to zero. If I is not equal to J or you I dotted with you. I is equal to one because we&#x27;re given that we have orthogonal matrices here. So this condition right here tells us that the dot product of any to different vectors in this matrix would be zero. So, for example, these are just some examples. And if two vectors orthogonal, that means that they do not lie on the same line. So these vectors air all linearly independent. So therefore this&#x27;ll set is linearly independent. So this forms a basis in orthogonal basis Also, and then we just need to prove that these vectors are also normal. So they have magnitude one. And to do that, the magnitude of use of I would just be you survived dotted with itself or the magnitude squared would just be one. So therefore, the magnitude of each of these vectors is also one. So it&#x27;s so it&#x27;s an orthogonal basis and a normal basis. So therefore, it&#x27;s a North a normal basis, and we&#x27;re done.

Donald Albin · Answer

a really n by n matrix A is cook called orthogonal. If a times a transpose is the same as a transpose times a which is the identity matrix. If a is an artha orthogonal matrix, then prove that the determine it of a is plus or minus one. Okay, well, the determine it, uh, the identity matrix is one. So that means And the reason is, if we have 1111 no matter how many ones there are, how much is years over here? And a whole bunch of zeroes over here is Multiply that down the diagonal and you get one. Okay, so that means that the determine it of a times a transpose is one, but the determinant of eight times a transpose. According to postulate, nine is the determine it of a times the determine it of a transpose. But according to postulate five, the determined of a determinant of a transposed is determined of a So if the determinant of a squared is one than the determinant of a could be plus or minus one because negative one squared is also one

Pam Russell · Answer

okay to show that Matrix A, which is co sign Alfa Sign Alfa Negative sign Alfa and Co sign Alfa is orthogonal. We&#x27;re going to need to show that the transpose of this matrix is equal to the inverse of this matrix. So we&#x27;ll start with the transpose. And so with the transpose of the matrix, the columns become the rose. And so here we will have signed Alfa coastline. Beautiful. Okay, now we need to find the inverse. So we&#x27;ll use the gallows Jordan method and we will enjoying the identity matrix, and then we&#x27;ll use our row operation. So I will divide row One by co sign and then replace it. And so I will get one sign Alfa over co sign Alfa One over co sign Alfa zero and then wrote to doesn&#x27;t change. Okay, so here I&#x27;m going to multiply row one by sign Alfa and add it to row two and then replace row two. And so we know that where one is not going to change. So I&#x27;m just gonna write it out, and so we&#x27;ll get sign out from minus sign Alfa. And then, in this case here, we&#x27;re going to have signed Alfa Times Sign Alfa over co sign Helpful plus Co sign fo So we get signed, squared over, co sign and then you have common denominators. We&#x27;ll be right that fraction. This is just equal to sine squared plus co sine squared Alfa all over coastline. Alfa Numerator is an identity and so we have one over co sign Alfa, Remember, The identity I&#x27;m talking about is that the sine squared Alfa plus Co. Sine squared Alfa is equal to one and then we&#x27;ll multiply, sign Alfa over co sign Alfa and added to zero. So we have signed Alfa over Coastline Alfa and then in this position we have warm All right, so let&#x27;s go ahead and multiply road to buy co sign Alfa and then replace it. So Row one won&#x27;t change and we get 01 Coastlines cancel and we get signed Alfa and then we get co sign of Alka. The last thing we need to do is turn this entry into zero. So that is tangent. Or we could leave it as sign over co sign. So let&#x27;s go ahead and multiply by the negative tangent or sign over co sign Times Road to added to row one and replace roll one. So we know that Row two isn&#x27;t changing. And so this is a one negative tangent plus tangent. That&#x27;s gonna be a zero. So now we have negative sign Alfa over co sign Alfa Times sign l phone plus one over co se No! So this is gonna be sine squared Alfa over co sign Alfa plus one over co sign Alfa and that&#x27;ll be a negative until we get one minus sine squared Alfa over co sign Alfa But this is an identity. So if we come back and look at our previous identity appear that we talked about If I rewrite this as co sine squared Alfa is equal to one minus sine squared Alfa that I can come down here and plug that in and I get co sine squared Alfa over cosign your Alfa which gives me co sign Alfa. So right here will have co sign Alfa and then were multiplying Negative sign Alfa over co sign Alfa Times co sign Alfa and that&#x27;s going to give me a sign. Helpful And so here is our inverse matrix And let&#x27;s compare that now to our transposed and it is identical. And so we&#x27;ve shown that the transposed is equal to the inverse and therefore this matrix is orthogonal.

Adhish Rele · Answer

in this example, we have been given a Matrix E, and we need to prove that is orthogonal. So let us review. What does our terminal meet tricks mean? So he is or talk if the transpose equals Klay Inverse. Okay, so we need to figure out what is the transport off given Matrix A and orders in words. So this was framed the transpose me tricks, since the it&#x27;s the easier one off the to. So it transpose is just or do do is the first column becomes the first true. And though second column becomes the second, the first column it was called him over a year is this. And it is Route 3/2 at the next element is negative. One work on that will better second element in the first group. The next column is 1/2 and with three by two. So that will be a second room one have on three over to Okay, So we now. So we just mark this And this is a second role in Transports, which is the first, which is a second column in a Okay, since we have figured out what e and a transpose are Let&#x27;s start tomorrow to play them a into a transpose is just equal toe. He is 3/2 negative. Well ah, with three over toe and he transposed is a tree or toe. One negro one have, uh, and three by two let&#x27;s apply matrix multiplication. This will just be with 3/2 multiplied by day or two. That&#x27;s three or four. Bless 1/4. Next Janet tree is negative. Three or four positive Route three or four. The next entry is negative. Route three or four positive through three or four, and the last one is 3/4. Plus, I&#x27;m sorry. The last entry, the Russian treat the force entry is 1/4 place three or four, so there&#x27;s just simplify stool 100 and one. Therefore you get a multi member e transposes I. Similarly, we can verify that e transposed repair by E is also equal to the identity matrix, and hence we can see that a transpose equals a inwardness. Therefore, we can conclude that a trans that e it is indeed is indeed or Taliban, and that&#x27;s the solution for this example

Pam Russell · Answer

to show that Matrix a 01 negative 10 is orthogonal. We&#x27;ll need to show that it&#x27;s transposed is equal to its inverse. So starting with the transpose so the transpose of a will be to take our column one and make that into Row One and column two will now become road to, So there&#x27;s are transposed. Now let&#x27;s go ahead and calculate the inverse so we use the girls Jordan method So 01 negative 10 1001 So let&#x27;s go ahead and switch row one and row two. And at the same time I will multiply row to buy a negative value. Negative one. And so we get 100 negative one 0110 And so we have our inverse matrix, and you can see that are inverse. Matrix is equal to the transposed, and so therefore, a is or is an Ortho agonal matrix

Robert Daugherty · Answer

Section 32 Problem 51. They give us a condition where the transpose of an MBA and matrix is equal. Toothy, inverse. Um and they say If this is make, this is what&#x27;s called a non orthogonal matrix proved that if this is a situation than the determined of a has to be plus or minus one. So by definition and orthogonal matrix satisfies this condition, properties of determinants would tell us that the determinant of the transpose is equal to the determine age of the inverse. Well, we also know that properties of determinants is a determinate of a is equal to the determinant of the transpose. So this would tell me that the determinant of a is equal to I also know that the determinant of a inverse does he go toe one over the determinant of a so that gives me this situation here won&#x27;t get the determinant of a is one over the determinant of a So if you multiply both sides by the determinant of a you get the determinant of a square, does it with one. Take the square root about signs and you get the determinant of a the secret of plus or minus quivered at one therefore, determinant of a is equal to plus or minus one

Problem

Let $U$ and $V$ be $n \times n$ orthogonal matric…

Problem 28 Easy Difficulty

Let $U$ be an $n \times n$ orthogonal matrix. Show that the rows of $U$ form an orthonormal basis of $\mathbb{R}^{n}$ .

Answer

This implies that, the rows of $U$ form an orthonormal basis for $\mathbb{R}^{n}$ .

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Heather Z.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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This implies that, the rows of $U$ form an orthonormal basis for $\mathbb{R}^{n}$ .

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Heather Z.

Joseph L.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Catherine R.

Anna Marie V.

Heather Z.

Joseph L.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications