Let n-order non pinchable matrix A and n-dimensional column vector α satisfy R {(first row) a α (second row) α t (transpose) 0} = R (a), then the equations AX = α must have infinite solutions Why?

Let n-order non pinchable matrix A and n-dimensional column vector α satisfy R {(first row) a α (second row) α t (transpose) 0} = R (a), then the equations AX = α must have infinite solutions Why?

It is known that α can be expressed linearly by the set of column vectors of A
So AX = α has a solution
And because a is irreversible, R (a)

10. Y is an orthogonal n-dimensional column vector. Note a = XY ^ T. why is the square of a equal to 0
10. Y is an orthogonal n-dimensional column vector
Let a = XY ^ T. why is the square of a equal to 0

The square of a is (XY ^ t) (XY ^ t),
The matrix satisfies the associative law (y ^ t, x) = 0, [this 0 is a number]
So the square of a is equal to 0

Let α be an n-dimensional column vector and E be an n-order identity matrix. It is proved that a = E-2 α ^ t / (α ^ t α) is an orthogonal matrix

Prove: because a = E-2 α ^ t / (α ^ t α), a ^ t = e ^ T-2 (α α ^ t) ^ t / (α ^ t α) = E-2 α ^ t / (α ^ t α), so AA ^ t = [E-2 α ^ t / (α ^ t α)] = E-2 α ^ t / (α ^ t α) - 2 α ^ t / (α ^ t α) + 4 α ^ t α ^ t α ^ t / (α ^ t α) ^ 2 = e -

XY is a mutually orthogonal n-dimensional column vector, which conclusion can be explained?

X·Y＝0,
10. The sum of products of corresponding components of y = 0
As a matrix product, X ′ y = y ′ x = 0 (zero matrix)