Is an invertible matrix a square matrix? So, must it be of the same order?

Is an invertible matrix a square matrix? So, must it be of the same order?

Linear algebra book definition: for n-order matrix A, if there is an n-order matrix B such that ab = Ba = e, then the matrix A is invertible. Under this concept, it must be a square matrix, and what we started to learn is only a square matrix. If you study deeply, consider the generalized inverse, then it can be m * n

Linear Algebra: does a square matrix belong to a matrix?

Of course! Square matrix, square matrix

How to find the nth power of matrix (m ^ n) * a
In the elective 4-2 matrix and transformation have to write, but the book is not around, trouble friends with books to teach

The multiplication of matrix is very complicated, let alone the nth power
The general matrix M is some special matrix. Either m multiplies itself or m itself. Or m becomes 0 after 2 ~ 3 power. In a word, it will be regular to test 2 ~ 3 power first

A is a 2x2 matrix. It is proved that if the K power of a is equal to zero and K is greater than 2, then the square of a is equal to 0

If you've learned eigenvalues, obviously the eigenvalue of a is 0, and then
Method 1: using Jordan standard model
Method 2: using Cayley Hamilton theorem
If you haven't learned eigenvalues
Method 3: the rank of a is no more than 1, so a = XY ^ t, where x and y are vectors of 2x1
A^k = (y^Tx)^{k-1} A = 0 => y^Tx=0 => A^2=0

Let a square matrix a satisfy a ^ 2-a-2e = 0. Prove that a and a + 2E are invertible, and find a ^ (- 1) and (a + 2e) ^ (- 1)
College homework, who can help^^

A(A-E)=2E
(A+2E)(A-3E)=-4E

Let the square matrix a satisfy a ^ 2-a-2e = 0, and prove that a and a + 2E are invertible

A^2-A-2E=0
A^2-A=2E
A(A-E)=2E
So a / 2 and (A-E) are inverse
In the same way
A^2-A-2E=0
A^2-A-6E=-4E
(A-3E)(A+2E)=-4E
Do you see the opposite?

It is proved that a square matrix a satisfies the relation aa-2a-2e = 0, a and a + 2E can be inversed, and the inverse matrix is obtained

Because a & sup2; - 2A-2E = a (a-2e) - 2E = 0
So a (a-2e) = 2E
A (1/2)(A-2E)=E
So a is reversible and a inverse is (1 / 2) (a-2e)
But a & sup2; - 2A-2E = (A-4E) (a + 2e) + 6e = 0
Similar to the previous discussion,
A + 2E is reversible and the inverse is (- 1 / 6) (A-4E)

Let a be a square matrix of order n, the fourth power of a - the second power of 5A + 4E = 0, and prove that a is invertible

Because a ^ 4-5a ^ 2 + 4E = 0
So a (a ^ 3-5a) = - 4E
So a is reversible and (- 1 / 4) (a ^ 3-5a)

Let a square matrix of order n satisfy the following condition: the square of a-a-2e = 0. It is proved that a and a + 2E are invertible and their inverses are obtained

Let a (A-E) = 2E, then the inverse of a is 1 / 2 (A-E)
Similarly, (a + 2e) (a-3e) = A & sup2; - a-6e = - 4E, so the inverse of (a + 2e) is - 1 / 4 (a-3e)

If the square matrix a satisfies the equation a square-2a + 3I = 0, then a and a-3i are all invertible, and their inverse matrices are obtained. How to prove?

Proof: because a ^ 2-2a + 3I = 0
So a (a-2i) = - 3I
So a is reversible and a ^ - 1 = (- 1 / 3) (a - 2I)
And a ^ 2-2a + 3I = 0
A (a-3i) + a-3i + 6I = 0
So (a-3i) (a + I) = - 6I
So a-3i is reversible and (a-3i) ^ - 1 = (- 1 / 6) (a + I)