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Can a Matrix with Zero Eigenvalue Be Invertible?
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[QUOTE="slider142, post: 5634833, member: 66092"] Hi mr-feeno, Invertible means the same thing as in "invertible function". We are seeking a "reverse" function that can give us back the original vector from the image of a vector in the range of the matrix transformation. That is, consider the matrix A as a function f(v) = Av, where Av is ordinary matrix multiplication of the matrix A with an arbitrary vector v. If Av = 0v = 0 for any particular vector v, notice what happens if we try any vector that is a scalar multiple of v as an input to our function. Suppose w = kv for any scalar k. Then f(w) = Aw = Akv = kAv = k*0 = 0. So every single vector that is a scalar multiple of v will be sent to 0 if v is sent to 0. Think of trying to invert this action. Given 0, we would like the inverse function to tell us which vector was sent to 0. Is it possible, given that more than one vector was sent to 0? If you need to do it with arithmetic, then we recall that a square matrix is not invertible if and only if its determinant is 0 (this theorem is part of a large laundry list of basic interrelated determinant properties that you should have somewhere in your text or notes). As you have shown that the determinant is 0, you may then imply that the square matrix is not invertible (also referred to as being "singular"). The reason why was described in greater detail above. For part 1., the fact that it is a 2x2 matrix that is symmetric means it must be of the form [tex]\begin{bmatrix}a & b\\ b & c\end{bmatrix}[/tex] for some numbers a, b, and c. Try to find the eigenvalues and eigenvectors of this type of matrix. What does this tell you? [/QUOTE]
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Can a Matrix with Zero Eigenvalue Be Invertible?
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