Let and

For which values of the parameter a is , and a basis for ?

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- November 4th 2009, 05:36 PMZockenBasis/Spans
Let and

For which values of the parameter a is , and a basis for ? - November 4th 2009, 05:39 PMBruno J.
Hint : evaluate the determinant of the matrix whose columns are those three vectors. Solve for the values of for which this determinant is nonzero. (Do you see why?)

- November 4th 2009, 05:52 PMZocken
Bruno, I found the determinant of the matrix to be and the values of a that make it nonzero are and

Is this right? And could you explain why this works (I am used to row reducing from class)? Thanks - November 4th 2009, 06:04 PMBruno J.
How can you possibly get in there? You should get a second degree polynomial in .

This works, because for the three vectors to be a basis of they should form the sides of a parallelepiped having nonzero volume, and the determinant is that volume. Or, more abstractly put, if is a basis, then the change of basis transformation from to the standard basis is invertible; its matrix representation is the one whose determinant you evaluated (with a small mistake). - November 4th 2009, 06:13 PMmath2009
Determinant is easy method. But when dimension is high, it's difficult to solve equation.

I suggest processing by**rref**

if three vectors span a basis, then rank(A)=3, - November 5th 2009, 12:22 AMZockenSpan
thank you