Knowing that a third order matrix is an orthogonal matrix, how to find one of the unknowns

Yesterday I was stuck with a problem for a long time, which was theoretically possible, but I was trying to write a better solution, so I thought about it for a while. And the background of it, which I later found out, was an orthogonal matrix.

The definition of an orthogonal matrix may seem like nothing special at a rough glance, which is the matrix [formula] that makes the [formula].

where [Formula] represents the transpose matrix of [Formula] and [Formula] represents the identity matrix.

We point out the algebraic properties of some orthogonal matrices:

1.The product [formula] of two [formula] level orthogonal matrices [formula] is also an orthogonal matrix. This is because.

Formula] 2.The identity matrix is an orthogonal matrix. This is obvious.

3.Multiplication according to the matrix.

, the orthogonal matrix must be reversible, and the determinant is [formula].

Further, it is true that [Formula] and [Formula] are also orthogonal matrices, and the determinant is equal to [Formula].

So the whole [formula] level orthogonal matrix with respect to the multiplication of the matrix is a group.

Then there are more magical properties, which require the introduction of Euclidean space.

Then explain the origin of the name of the orthogonal matrix.

Euclidean space is first and foremost [formula] and second is ordinary linear space.

Formula] is the addition of an inner product.

Operation: Formula].

Calling [formula] orthogonal refers to [formula].

In addition, with the inner product it is possible to define the length. Formula].

If you think of a third-order matrix as three column vectors, then the three column vectors are orthogonal in pairs, and the modulo length is 1.

Two pairs orthogonal means that the inner product is 0, for example, there are ( 2 2 (1,-1,0)t,( 3 3)(-1,1,1)t,(1 (2 +a ))2,a,0)t

Well, the coefficients that can be ignored for the unitization have two equations:

1*2+(-1)*a+0*0=2-a=0

1*2+1*a+0*0=a-2=0

then a=2(Generally speaking, an unknown can be solved by one equation) but there are also cases that cannot be solved orthogonally, for example: you decompose into (1,0,0)t, (0,1,0)t, (0,0,a)t

Then there are two equations.

1*0+0*0+0*a=0

0*0+1*0+0*a=0

Apparently it's all true.

Then we use the property of the modulus length of 1, 0+0+a=1 to get a = 1

The column rank is equal to 2

One column can be represented linearly by the remaining two.

For example, a1= k2a2+k3a3

Then c1 - k2c2 - k3c3

The first column is all reduced to 0.

So the determinant.

Equal to 0 also cavity round bi can be directly from the rank of the matrix.

The definition of see. The rank of the matrix is the order of the highest order non-zero.

A rank of 2 and a 3rd order sub is equal to 0

The determinant of an orthogonal matrix is equal to 1 or -1

So |a*| a|^(9-1) =a|^8 = 1.

Column rank is equal to 2

One column can be represented linearly by the remaining two.

For example, a1= k2a2+k3a3

Then c1 - k2c2 - k3c3

The first column is all reduced to 0.

So the determinant is equal to 0

It can also be seen directly from the definition of the rank of the matrix.

The rank of the matrix is the order of the highest order non-zero.

A rank of 2 and a 3rd order sub is equal to 0

The rank of the 3rd order square matrix is 2, which means that there are two coincidental formulas of the modified square, 1x1 and 2x2, so there must be a row or column of elements in the square matrix that are all 0

The so-called triangular matrix on the silver hold, that is, a matrix a=(aij), when i>j, there is aij=0The evidence of this topic is as follows: Evidence of grinding potatoes:

Let a=(aij) and b=(bij) be the nth order square of the upper triangle, then >aij=bij=0note c = ab = cij) then cij = ai1b1j+.. when i>jaii-1bi-1j + ai,ibi,..

There are many such matrices, and the following diagram is two examples. The Economic Mathematics team will help you solve the problem, please adopt it in time. Thank you!

Let the matrix sought be a, the unknown quantity in a is x, and the diagonal matrix similar to a is b

1) If x is on the diagonal of a, then tr(a)=tr(b) can directly find x (similar matrix traces are equal);

2) x is anywhere in a, then by |a|=|b|x (the determinant of the similarity matrix is equal) can also be found.