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For a square matrix (as your example is), the rank is full if and only if the determinant is nonzero 2) if it has n distinct eigenvalues its rank is atleast n. When there are zeros in nice positions of the matrix, it can be easier to calculate the determinant (so it is in this case).
The first two colums are linearly independent so the rank is $2$. 1) if a matrix has 1 eigenvalue as zero, the dimension of its kernel may be 1 or more (depends upon the number of other eigenvalues) From there, you can get the rank from the rank theorem
But we do have an upper bound, the dimension of each eigenspace (geometric multiplicity) cannot be larger than the (algebraic) multiplicity of the eigenvalue in general
By a theorem that i've studied it the row rank and the column rank of a matrix are same But the book wants the column rank of the given matrix by calculation and i can't find out it column rank. It is often taken as the definition of rank of a matrix I see a proof of the determinant rank being the same as the row rank in the book elementary linear algebra by kenneth kuttler, which i see in google books.
The rank of the matrix is equal to the number of nonzero rows in the matrix after reducing it to the row echelon form using elementary transformations over the rows of the matrix. From linear algebra we know that the rank of a matrix is the maximal number of linearly independent columns or rows in a matrix So, for a matrix, the rank can be determined by simple row reduction, I have some discussion with my friend about matrix rank
But we find that even we know how to compute rank, we don't know how to show the matrix is full rank
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