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randommacuser
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How do I prove that if an nxn matrix A is diagonalizable (is invertible, has rank n, etc.), its columns span Rn?
Ohh, yeahHurkyl said:A matrix doesn't have to be invertible to be diagonalizable...
thismatt grime said:Why can't the determinant of a Diagonal matrix be zero?
EvLer said:this
although I see your point, because eigenvalue can be 0, while eigenvector cannot. I am still confused though, because it is defined exactly the same way in my text-book as well, i.e. Q has to be invertible.
Was working on conjecture as Data suggested. Came to same conclusion (yooo-hooo, it worked!)matt grime said:So what, why is this confusing, what's that link got to do with what I wrote? At no point does it state either A or D are invertible, indeed A is invertible if and only if D is.
An invertible matrix is a square matrix that has a unique inverse matrix, meaning it can be multiplied with another matrix to give the identity matrix. The inverse of a matrix is denoted as A^{-1}.
To prove that a matrix is invertible, you can use various methods such as the determinant method, the row reduction method, or the adjugate method. These methods involve manipulating the matrix to show that it has an inverse.
No, not every square matrix can be inverted. A square matrix can only be inverted if its determinant is non-zero. If the determinant is zero, the matrix is said to be singular and does not have an inverse.
Invertible matrices are important in various mathematical applications, such as solving systems of equations, finding the inverse of a linear transformation, and diagonalizing matrices. They also have practical applications in fields such as engineering, computer science, and economics.
Yes, the inverse of an invertible matrix is unique. This means that there is only one possible inverse matrix for a given invertible matrix. It is also important to note that the inverse of a matrix is not the same as its transpose.